 let me just give you an overview. What are the conics that we are going to? Okay, conics or conic sections. So what are the conic sections that we are going to talk about under this particular chapter? The first chapter that we are going to start today itself is circles. Okay, circles will take one class. That means in today's session, I should be able to finish off circles from your school point of view. Again, I'm repeating from your school point of view, not from your J point of view. J point of view will take at least three more classes just for circles. Yeah, so you must be wondering, so big is the circle chapter so big for J. Yes, it is quite a big chapter. And from circles chapter, a lot of concepts will be taken as it is in your other conics like parabola ellipse and hyperbola. So there are certain things that we will derive in circle and those things will be taken as it is, okay, in your concepts of parabola hyperbola etc. Next is your, of course, parabola chapter, which you have already seen in your school ellipse. Okay, hyperbola, fine. And the fifth one is called a degenerate conic, which we call as the pair of straight lines, which we call as the pair of straight lines. Right. Now this is not there in CBC. It's not for CBC. Okay, if you are preparing for just school level exam, this topic has not been included in the CBC exam. It is only there for your competitive level exam. So all these topics, we are going to cover, okay, only for school exam point of view, as of now. But later on, when we revisit these chapters, we will be doing it to a very, very large depth. Okay. This board, not presently right now, I'm going to share my screen only and do it. So right now we are sitting in the center office. Okay, I'm not taking it from home. Okay, that is why you are seeing these boards. But yes, when we have this offline classes and all, we call students to this setup. Okay. Anyway, so let's get started. So let's start with circles. Very soon, we will keep you updated Arvind regarding that. Okay. So circles, first of all, let us understand the locus definition of a circle. What is the locus definition? Locus is something which you have already come across in straight lines. What is locus? Locus is a path. Locus is nothing but a path traced by a moving point, satisfying certain conditions, isn't it? So locus is nothing but a path traced by a moving point, satisfying certain condition or conditions. In the same way, if I have to write a locus condition for a circle, what would you say? So let us, you know, write this down, let us note this down. A circle is nothing but a path traced by a moving point, traced by a moving point, moving point. Okay, moving in a plane, moving in a plane, moving in a plane in such a way, in such a way that its distance from, its distance from a fixed point is always a constant. Distance from a fixed point is always a constant. Always a constant. Okay, so just to show you how this is work, so let us say this is a fixed point. So if this point moves, this is your fixed point called C and if there is a point P, which is a moving point, which moves in such a way that this distance is always a constant. Okay, let's say I call this distance as R. Okay, so if this distance is always R, then the path traced by this moving point will be that of a circle. So let me just draw it like this. So this will be the path traced by that moving point. Is it fine? Now we already know from our class 10th and of course to a certain extent in class 9th also we have done that this fixed point is called the center of the circle. Okay, this fixed point is called the center of the circle and this fixed distance or this constant distance is what we call as the radius of the circle. This is called the radius of the circle. Okay, now based on this definition, we will be deriving the equation of this circle. We will be finding the locus equation of the circle. So let's say this is alpha, beta and this is h comma q. Okay, if I write down the locus definition, this says Pa should be equal to R. And let us use our distance formula. So Pa, I'm sorry, Pa would be nothing but under root of h minus alpha whole square k minus beta whole square equal to R. So just square both the sides, let's square both the sides, square both sides and this will lead to h minus alpha the whole square k minus beta the whole square equal to R square. And now we generalize it. Let us generalize it by putting our h as an x and k as a y. So when we do that, we end up getting the equation as x minus alpha the whole square y minus beta the whole square equal to R square. This equation of the circle is known as the center radius form of the equation of a circle. So we call this as the center radius form of the equation of a circle. Because when we derived this equation, we were given the center coordinates and the radius value of the circle. Okay, so everybody please make a note of this very, very important. And then we'll do some analysis of this particular equation in some time. Okay, any questions, any concerns so far, please do let me know. Okay, everybody's fine. Let's take a simple example. Just one simple question we'll take up. Let's take this question. Find the find the equation of a circle, equation of a circle with center at the center at 2 comma 3 and radius as five units, radius equal to five units. Okay, so everybody please do this just you know, basic, basic beginning level question on this particular topic. So you know the center, center is at 2 comma 3 and radius is given to us five units. What is the equation of the circle? Just write it done on the chat box once you're done. Very good. So most of you already got it. This is x minus 2 the whole square y minus 3 the whole square equal to five square. Okay, now if you simplify this, if you simplify this, the result is going to look like this. The result is going to look like this. Again, let's further simplify this. Okay, and this is going to be 13 minus 25, which is minus 12 equal to zero. Okay, now if you look at the equation of the circle, okay, you would realize that a question of such a nature looks like something of this nature, 2 gx, 2 fi plus c equal to zero. Okay, so this is the general form of the equation of a circle. This is the general form of the equation of the circle. Don't get confused. The previous form was called the center radius form. I'll go back to the previous screen. This is called the center radius form. And now what I have written on your screen here, this is called the general form of the equation of a circle. So if I use this word, like you know, write down the general form, you should automatically understand that sir is asking us to write the equation in this form. Here, please note that the coefficient of x square and the coefficient of y square both are equal to one each. Both are equal to one each. Please note this down. That means if they are not one, I cannot call it as a general form. Okay, please note this down. If the coefficient of x square and y square is not one, we cannot claim that equation to be the general form of the equation. Is it clear? Okay, now I'll be spending some time on this particular form just to give you something very important over here. So number one, if you compare this, if you compare this equation with our center radius form, what was our center radius form x minus alpha the whole square y minus beta the whole square equal to r square. So you would realize that you get x square plus y square minus 2 x alpha minus 2 y beta plus alpha square beta square minus r square equal to zero. I've just expanded this for you all. Okay, now compare one and two. So if you compare one and two, let me call this as one and let me call this as two. Compare one and two. So if you compare one and two, you would realize that 2g is minus 2 alpha. That means alpha is minus g and 2f is minus 2 beta. That means beta is minus f. Okay, and not only that, the term c, the term c that you have over here, that will be this term. This is your c and this is the constant here. So if you compare them, you will end up getting c is equal to alpha square plus beta square minus r square is equal to your c. Okay, so please note these three, please note these d. From here, we get g as or you can say here we get the center as alpha beta. Alpha beta is nothing but minus g minus f. Now, why did I give you this relation is because many a times you will be given the equation of a circle in the general form. So from that general form, if you want to know what is the center of that given circle, this is what you need to write down for the center. Okay, so this is to be kept in mind because many questions will be framed in your general form. So from your general form, if you need to know the center, this is what you are going to use. Okay, now from the last equation that you see, since your alpha is minus g, you can write here like this. Let me write it g square f square beta is minus f correct. So beta square will be minus f whole square minus r square equal to c. In other words, r square is g square plus f square minus c. In short, your r is going to become, your radius is going to become under root of g square plus f square minus c. Okay, so these two aspects, please note this down. They're very, very important because you will be needing this in almost every question where you are given with a general form. So from general form, if somebody wants to get the center, this is what you are going to use. And if you want to get the radius, this is what you're going to use. Now a very important point to be noted here that this formula is only going to work if your equation of the circle is in the general form. That means coefficient of x square and y square has been made one. Please note this down. So this, I will write it down also. So please note that these two formulae, they will work only for the general form of the equation, form of the equation. Are you getting my point? That means if you have been given some other form, which is not a general form, in other words, the coefficient of x square and y square are not one, then this formula is not going to work. Okay, that's a very common mistake that people normally do in the beginning part of this chapter. So if in UT or semester exams, you get such questions, be careful before using this formula. Ask yourself not once but twice. Is my equation given to me in general form? Is my coefficient of x square and y square has been made one? If the answer to that is yes, then only you can use this to get your center and radius, else not. Is it fine? That's number one analysis that I want to present here. Number two, please note this down. This is number one, number two, number three, et cetera. I'll talk about it in some time. The second important thing that I would like to highlight here, can I go to the next slide if everybody has copied? Yes, sure, scroll to the right. Scroll to the right means scroll to this side. This is what you want to see. September, is this fine? The other side, scroll to the right. This is the place? Okay, great. Anybody else who would like to copy here? Okay. Next thing that we would like to understand here, that is the second point is in any circle, please note down. In any circle, the coefficient of x square and the coefficient of y square must be equal. Whether they are one or not, but they should be equal. If it is one, then it has been mentioned in a standard form. But normally what happens is people multiply throughout with some constant. Let's say they multiply throughout with a three. But whatever you do in the equation of a circle, this criteria must definitely be satisfied. So if the conic section represents the equation of a circle, then in that equation, your coefficient of x square and y square must be equal. By the way, y-serversa is not true. That means if the coefficient of x square and y square are both equal in any conic section equation, it need not represent a circle. But if it is a circle, then this condition must be satisfied. As you can see here also in our expression of general form or center radius form, if you expand it, your coefficient of x square and y square will be both be equal to each other. Of course, it is one in this case, but what if I multiply throughout with a 10 or a minus 5? Still it will be equal to each other. So that is number two observation, which anyways you would have already got from your school as well. The third observation that you must note down that there is no xy term in the equation of a circle. There is no xy term or xy related term in the equation of the circle, in the equation of a circle. So a circle equation will never contain an xy term. Again, y-serversa is not true or y-serversa need not be true. That means if there is no xy term, that doesn't mean it is definitely a circle. No, I don't mean to say that. If there is a circle equation, it must definitely not have xy. But an expression or an equation or a second degree equation that you have, if at all it has an xy term, you can say for sure it is not a circle. Are you getting my point? So here is again the summary. If there is an xy term, it cannot represent a circle. It will represent something else, some other conic, maybe a parabola, ellipse, hyperbola, but not a circle for sure. And also please note that if there is no xy term, it need not always represent a circle. Are you getting my point? Are you getting the differences in the way I am speaking them out? So one thing for sure, a circle equation will never contain xy term and in that the coefficient of x square and the coefficient of y square will always be equal. And of course, if you have been given the equation x square, y square, 2 gx, 2 fy plus c equal to 0, for it to represent a real circle, its radius must be greater than equal to 0. So for this equation to represent a real circle, real circle is something which you can actually draw on the Cartesian plane. So in order to draw a real circle, the radius must be greater than equal to 0. And what is the radius expression we just now found out? Radius expression was something like this. So this must be greater than equal to 0. That means your g square plus f square minus c must be greater than equal to 0. Please note this down. So please note down that in the equation of a real circle, g square plus f square minus c should be greater than equal to 0. So should we summarize this? So x square plus y square plus 2 gx plus 2 fy plus c. Okay, let me summarize it over here. Let me write a summary. So let's write a summary here. So the general form, this is the center radius form. And this is the general form. I will write down both of them. Okay, this is called the general form. Center radius form is x minus alpha the whole square y minus beta the whole square equal to r square. And general form is x square plus y square plus 2 gx plus 2 fy plus c equal to 0. From the general form, the center is found out by using the formula minus g minus f. Okay, please note that in the general form, the coefficient of x square and y square both are one each. Okay, let me not write that down. And the radius is given by under root of g square plus f square minus c. Okay, few things to be noted here. Number one, the coefficient of x square, y square should be equal, whether one or not one. Okay, and if it is given to you in the general form, then the radius must be greater than equal to zero for you to have a real circle, for you to have a real circle. Is there any question with respect to this whatever we have discussed so far? Any questions, any concerns? Please feel free to raise the concern so that we can discuss it out. All right, no issues. Let's take a question. Let's take a question. Let's begin with this question. Find the center and the radius of the circle. Please give me a response on the chat box. Yes. Okay, very good, Neil. Anybody else? I'm assuming you people have done basic questions on circles in your school as well, right? So these questions should not look new to you, right? Yeah. Okay, so Neil has given one response. Ashinthia has given Neil's response and Ashinthia's response. However, are different. Okay, let's wait for others. Okay, okay, Jatin, Arya, Vishal can be great. Should we discuss it? Easy question. See, first of all, when you look at this equation, okay, let's bring everything to one side. Now answer this simple question of mine. Is this in the general form? Is this equation of the circle in the general form? Just answer with a yes or a no. No, right? This is not a general form. Okay, the reason for it not being a general form is that you have a 2 in the coefficient of x square and y square. However, to find the general form or to write the general form, we must have the coefficient as 1 each. So this is first of all, what we need to do. We need to divide throughout with a 2. Okay, and this is how the equation will look like. Now here, if you compare, this will be your 2g. Okay, so 2g is minus 3 by 2. And this will be your 2f. So 2f will be 5 by 2. So which means g is negative 3 by 4. And sorry, this is f. I wrote a g again. Sorry for that typo. Okay, so this will means f is equal to 5 by 4. So where is the center for such a circle? Center for such a circle is minus g minus f. So your answer is going to be 3 by 4 comma minus 5 by 4. Okay, this is going to be the center. Now what is going to be the radius? Radius I've already discussed is g square plus f square minus c. So this answer is going to be 9 by 16, that is 3 by 4 the whole square, 25 by 16. And minus c means 7 by 2. Okay, minus of minus 7 by 2. So we can do one simple step over here. Let's write everything to the base of 16. So this will become 9, 25 and this will become 56, 56 by 16. So this is nothing but under root of 90. 34 plus 56 is 90. 90 I think you can write it as 3 by 4 root 10. Okay, so this will become your radius of the given circle. Is it fine? Any questions? Let me check how many of you got this right actually. I think Achintya was the only person who got everything right. Rest everybody has made some or the other mistakes. Okay, is it fine? Now your mistake is clear of what all mistake you are making. So such kind of equations may be given to you to deceive you where they will not write it in the general form. They will multiply this entire equation with something. Okay, and they will just ask you for the center and the radius. So before you go for center and the radius, ensure your circle has been made in the general form. That means coefficient of x square and y square have been made one each. Got the point? All right, let's take more questions. This was supposed to be a very simple warm-up question. Let's take this question, prove that the radii of the circle, this and this and this. So there are three circles here. So this is one circle, this is another circle, and this is another circle. So prove that the radii of these three circles are in arithmetic progression. Just say done on the chat box once you're done with it. Done, very good. Okay, so should we discuss it? So now if you look at the first equation, okay, as you can see the radius here, the radius here will be g square plus f square minus c under root. But there is no g, there is no f and c is minus one. Right, so it's zero, zero minus minus one under root. So this is a circle which we normally call as a unit circle. Okay, so this is a circle whose center is at origin and radius is one. Okay, so this is our unit circle. By the way, something very important which I would like to tell you. We already use the word center radius form, right? Okay, in that center radius form, if your center happens to be at the origin, then you end up getting the equation of a circle like this, correct? Such form of the equation of a circle is called the standard form. Okay, such form of the equation of a circle is called the standard form. So please note that we are using different types of names to call these equations. Center radius form, within the center radius form, if the center is at origin, we call it as the standard form. If you have x square, y square, 2gx, 2f5 plus c equal to zero, we call it as the general form. Okay, later on in today's session, you'll learn something called the diametrical form. You'll have something called the parametric form. So there are different types of the equations available. Okay, so now this is a circle whose radius is one. Okay, let's talk about the second one now. So second one, let us write in general form right now. So let's bring the 6 to the left side and let us use our formula under root of g square plus f square minus c, which is, by the way, what is g here? What is g value here? This is 2g. So g value is minus 1. This is 2f. So f value is minus 3. So this is 1 square plus 3 square minus minus 6. So this happens to be under root of 16, which is nothing but 4 units. Okay, let's look into the third circle as well. So third circle, let me use a different color marker. Maybe I can use a blue one. So for the third one, yeah, third one, the equation is x square plus y square minus 4x minus 12y. x square plus y square minus 12x minus 4y minus 9 equal to zero. Okay, so here g is minus 6, f is minus 2, c is minus 9. So if you use the formula, your radius will be under root of g square f square, let's write 2 square only, y to write minus 2 unnecessarily, minus minus 9. So that is nothing but 36 plus 4, 49 under root 49, which is 7 units. So yes, 1, 4, 7 is in arithmetic progression with a common difference of 3. So these are in arithmetic progression, hence proved. Is it fine? Any questions? Easy question? Anybody having any doubts? So this question was just to make you practice the expression for radius, etc. Because many people keep making mistakes, you know, related to the expression for the center, the expression for the radius. So this problem was like a good practice for you all. Is it fine? Any questions? Should we take the next one now? Okay, let's take the next one now. Okay, so let's take few questions where you would be asked for the equation of a circle where the direct information about radius and the center might not be given to you. Such questions will be very, very commonly asked in school exams as well. Okay, so here's a question almost of the similar type where the question sitter says, find the equation of a circle whose center is the point of intersection of these two lines and passes through the origin. Please give me your answer in the general form. Okay, so please type out your answer in the general form. Okay, it's a good practice. I mean, I would like to share this with you that whenever you write down the equation of certain, you know, curves, maybe a straight line, maybe a circle, etc., you must always try to present your answer in a general form. So leave your answer in the general form. Okay, okay, Harshita. So this question becomes actually very easy when you look at the entire situation. Let's say I sketch this diagram. Okay, roughly, let's say this is my two lines, which basically intersect. Okay, so let's say these are my two lines. Okay, so they intersect at the center and this circle is also passing through the origin. The circle is also passing through the origin and we need to get the equation of the circle. Okay, so let's say this is our radius. So very simple indirectly, they have given you the information to find the center. You just have to simultaneously solve these two straight lines equation. So once you simultaneously solve them, you end up getting your center of that given circle. So let's solve it simultaneously. Maybe let's multiply this with a three, multiply this with a two. Okay, let's rewrite it once again. So 6x minus 9y plus 12 equal to zero. This is 6x plus 8y minus 10 equal to zero. Subtract the results. Okay, subtract the results. So you will end up getting minus 17y plus 22 equal to zero. So y is equal to 22 upon 17. Okay, y is equal to 22 upon 17. Correct? Yes or no? People who have solved this, please rectify this result. Are you getting y value as 22 by 17 or something else? Because it looks so ugly, you know, it's hard to believe that you can have this answer as well. But okay, we have to take that as a pinch of salt. Maybe that's the answer for the y value. Okay, so let's find out our x. So 2x is equal to 3y minus 4. So 3y is 66 by 17 minus 4. That makes it minus 2 by 17. So x value comes out to be minus 1 by 17. Is this what you all are getting? Please confirm this. Okay, so this makes the center of the circle as minus 1 by 17 comma 22 by 17. Okay, now what information has been provided to us to get the radius? So the question center also says that the center, the circle passes through the origin. So the distance, the distance of the center from that origin will give you the radius. So radius, if I'm not mistaken, will be under root of 1 by 17 square plus 22 by 17 square. Yes or no? If I'm not mistaken, this comes out to be root 485 by 17. Okay, now once we know the center and we know the radius, can I not write down the center radius form? So let us write down the center radius form, center radius form, then we'll convert it back to the general form, not to worry. So it'll become x plus 1 by 17 the whole square, y minus 22 by 17 the whole square is equal to 485 by 17, that is square of r. Okay, let's multiply with 17 square throughout. Okay, let's multiply with 17. Let's do one thing. Let's open this expression. So this is, okay, this is this, this is, this is this, okay, and this is this. Now, we already can see, we already can see that this term is going to get cancelled off. This is 17 square. Yeah, this term, this term, and this term are going to get cancelled off. Correct? Multiply throughout with a 17. If you multiply throughout with a 17, you end up getting 17 x square to x 17 y square minus 44 y equal to zero. Is this fine? Any questions? Any concerns? Is this the result that you are getting or are you getting something else? Yes or no? Clear? Which part? Which part? This part? Where I'm finding the radius or the center? Which part do you want me to scroll up? Center. Okay, so center, I just solve this simultaneously, Arya. You can also do it at your own end. You need not follow exactly whatever I have written. Solving for the center means simultaneously solving these two linear equations. That's it. Okay. Okay. And then after this, I just expanded it the normal way. Okay, I cancelled out the terms which were getting cancelled off. And I just multiply it throughout with 17. That's it. Fine. No questions. All set. Okay. Let's take another one. Okay. Find the equation of a circle concentric with this and passing through minus two minus seven. So basically some indirect information has been provided to us to get the center and the radius. Please take the clue from the given information and get the equation of the circle. Again, just writing a done on the chat box is more than enough for me. You need not type out your answer. So just say a done on the chat box if you're done. Everybody knows the meaning of concentric. Concentric means having the same center as the given circle. Okay. So concentric means having the same center. Okay. So two circles are said to be concentric when they share the same center. Okay. Jatin is already done. Very good. Jatin. Okay. Neil also is done. Very good. Neil. Okay. Satyam, thank you for typing in the answer. Very good. We'll check in sometime. Okay. Let's discuss this. See, many of you would have spent some time finding the center. Okay. And you would have also spent some time finding the radius. Okay. And then you would have again written it as a center radius form and again converted it back to the general form. Right. But here is a simple trick to solve this question. Once you know that two circles are concentric, their general form will only differ in the constant. Okay. So it is like saying two lines are parallel. Even two lines are parallel. They only differ in their constants. Right. The coefficient of X and Y can be made the same for both the lines. Okay. In the same way, if the two equations are concentric, you can directly assume that let the equation of the circle that we need is this. Okay. Now only see something which we need to figure out. And that's an easy way to do it because you already know that the circle is passing through minus 2, minus 7. So substitute this point in place of X and Y and solve for C. So when you do that, you get 4 plus 49 plus 16 minus 42 plus C equal to 0. Yes or no. That's nothing but if I'm not mistaken, this is 53. 53 plus 16 is 69. 69 minus 42. 69 minus 42 is 27. Okay. So C value is going to be minus 27. Yes or no. Any questions? Correct. So just put the value of C here as minus 27. And our job is done. So the required equation of the circle will be X square plus Y square minus 8X plus 6Y minus 27 equal to 0. Is it fine? Very easier, simpler way to do it. Right. I'm sure most of you would have spent some time finding the center, which is, I mean, not a difficult ask. It is just 4 comma minus 3. But you have to spend some time finding the radius by using your distance formula. And then you would have written a center radius form again, simplified it and got the same result. But isn't this a simpler way to do it? Okay. Any questions? Any concerns? Any questions? Any concerns? Clear? Okay. Let's take another one. Let's take another one. Let's take this question. Okay. Let's take this locus question, prove that the locus of the center of the circle, half X square plus Y square plus X cos theta plus Y sin theta minus 4 equal to 0 is this. Now, before you start solving this question, first of all, everybody read this question carefully once. Now, let me ask you this question. Is this a fixed circle or is this circle dancing around? And if it is dancing around, why at all it is dancing around? How do you know it's a circle, which is a variable circle? Of course, you'll say, sir, the question is asking us to find the locus of the center that itself means it's a variable circle. That is fine. That is a lateral thinking. But what in this particular expression tells you it's a variable circle, the theta part, exactly. So the presence of this theta tells you that there is something happening to the center of the circle. Okay. There is something happening to the center of the circle. And because of that, the circle is a variable circle. And of course, because of theta, the center is moving. And if something is moving, it is going to move on a path. And you want to know the equation of the path it takes. That is why the whole question is made you to find the locus of the center of the circle. Please do this and just let me know once you're done. Oh, I thought this question is going to be super, super easy. Locust situation, locus questions, you have already done. Yeah, Harshita is done. Okay, very good. How about this? Okay, let us first write down this equation in a general form. Right now, it is not in the general form. So in the general form, this is the equation of the circle. Okay, let me ask you this question. What is the expression or what is the coordinates of the center of the circle? Look at this equation and tell me what is the coordinates of the center of the circle? So it says here 2g is 2 cos theta. Right. So minus g will be minus cos theta. Okay, and minus f will be minus sin theta. Now, this is like the h coordinate of the moving point and this is like the k coordinate of the moving point. Now, normally in order to find the locus, what do you do? You strike a relationship between h and k thereby eliminating the parameter. Right. That's what we do. Isn't it? So the simple way here to eliminate the parameter is square them and add them. Correct. Just square. Sorry. Yeah, just square them and add them. If you square them and add them, what do you get? You get a 1 and just generalize it. So when you generalize it, this becomes your equation of the locus and that's what the question center has asked us, proved. Is it fine? Is this the rocket science norm? So I think most of you are still scared of locus questions. Locus is not difficult at all but yes, you have to have a good command on how to solve locus questions. Okay. It's a very, very, you can say commonly asked subject matter in competitive exams. So you have to get good hold on finding locus. Okay. All right. So with this, we are now moving on to the next form of the equation of a circle, which we call as the diametrical form, diametrical form of the equation. Now again, we are not going to learn a different concept altogether. It's just a situation based form. Right. So if there is a center and the radius, you normally go for a center radius equation, right? So it's a situation based form. So depending upon the situation, you choose that form to write down the equation. In a similar way, if there is a situation where the question center has given you the diametrically opposite ends of a circle. Okay. Let's say this is a diametrically opposite ends of a circle, A, B. Okay. Let's say A is X1, Y1 and B is X2, Y2. And then the question center asks you what is the equation for such a circle? Now, I know many of you would be thinking, sir, big deal. We'll find the center by using midpoint. You already know A and B coordinates. So midpoint formula we will use to get the center. And just half the distance between A and B that will give you the radius. And then we know the center and the radius. So we are going to write the center radius form. Okay. Accepted. Very good. This approach will work. No doubt about it. But let's realize that there is an alternative way also to crack this situation or to find the equation of circle in this case. And what is that? All of you would realize that you can actually build up a locus condition here. And the locus condition is I can say that the circle would be the locus of all such points such that this angle is a 90 degree. This angle is a 90 degree. Am I right? So can I say this yellow circle is the locus of all such points such that at that point A, B subtends a right angle. Isn't it not? Take any point on this circle. Of course, you'll say what about A and B? In that case, it is a limiting case. Okay. So please understand at any point on this circle, at any point on this circle, A, B is going to subtend a right angle. Okay. Now I'll be using this definition to get the equation of this circle. See how. So if this is right angle, I can say the slope of AP into slope of PB, a slope of BP should be multiplying to give you a negative one. Correct. So what is the slope of AP? So slope of AP will be k minus y1. Okay divided by h minus x1 into slope of BP. Slope of BP is yk minus y2 by h minus x2. Yes or no? Now let's multiply this. Let's multiply this and take it to the other side. So you'll have negative of h minus x1, h minus x2. Okay. Bring everything to one side. If you bring everything to one side, this is how it will appear. Okay. Any doubt, any concerns till this stage? Do let me know. Everybody is happy till this stage. Okay. Everybody knows the slope concept. So product of the slopes is minus one. Now here we generalize. Let's generalize. So we generalize by putting our h as an x, k as a y. So when you do that, it becomes x minus x1, x minus x2 plus y minus y1, y minus y2 equal to zero. Right. This is a much convenient and a faster means to get the equation of the circle. So please note this down. This equation is called the diametrical form of the equation of the circle. So all you need to do is just use this formula. There is no need to find the center. There's no need to find the radius. All you need to do is just plug your extreme ends of your diameter into this expression and you are done. Okay. Kindly note this down. Is it fine? Any questions? Okay. Let's take problems based on the same. So let's take up few problems based on the concept. Let's start with find the equation of the circle, the end points of whose diameter are the centers of these two circles. Very simple formula based question. You may just write it down on the chat box. No need to give your answer. Just say I'm done. That's more than enough for me. Okay. Well, okay. So let's just figure out the center. So as per the given question, as per the given question, there are two circles and the centers of those two circles, let me draw two miniature circles over here. Yeah. The center of these two circles, they serve as the diametrically opposite ends or the ends of the diameter of our required circle. So let's say this green circle is our required circle. Let me make it smaller, smaller, smaller, smaller, smaller, smaller. Yeah. Yeah. So now we need to know first of all, what is the center of these two small red circles? So center we already know. So for the first one, the center would be, let's say this is our first circle. This is our second circle. Okay. This is our first one. This is our second one. So for the first one, the center is at minus three comma seven. Am I right? Isn't it? I hope by this time, everybody knows how to find center from the general form, correct? Convinced. Okay. Now what about this one? This one is going to be two comma minus five, two comma minus five. So once you know the extreme ends of the diameter, you can use the formula. I'll write it down once again for you. The diametrical form is x minus x1, x minus x2 plus y minus y1, y minus y2 equal to zero. So that's nothing but x plus three, x minus two plus plus y minus seven, y plus five equal to zero. Okay. So let's simplify this. So this is going to be x square. From here, I'll get y square. This is going to give us an x. This is going to give us a minus two y. And I think the constant term will be negative six minus 35, which is 41 equal to zero. Is it fine? Is this what you're all getting? Okay. So this becomes your desired equation of the circle. Any questions, any concerns, please do highlight. Clear? So there's absolutely no need to find the center. There's absolutely no need to find the radius. And then there's absolutely no need to invest any time in finding the center radius form and then converting it back to the general form. This does the work automatically for you. Okay. Let's take this one. The sides of a square are given to you. Find the equation of the circle drawn on the diagonals of the square as its diameter. Okay. Please sketch the graph. Please make a rough sketch of the scenario. Done. Very good. So let's try to sketch this in da scenario. So you have been given that there is a square which is formed by these lines. By the way, if you look at these lines, they actually represent vertical lines. For example, x equal to two and x equal to three, these are vertical lines. Okay. y equal to one and y equal to two, these are horizontal lines. Okay. So you end up forming a square over here. So this is your desired square. Okay. Now, the diagonal of the square as a diameter, you are sketching a circle. Okay, you're sketching a circle. Let me use a different color. Let's use a yellow color. Okay, you're sketching a circle. So what are the equations of this circle? Now, if this diagonal is a diameter, so would be this diagonal. Now it is up to you to use any one of the diagonals or any one of them as a diametrically opposite points. Okay. So let me name it. You could either use AD, sorry, AC or BD as your diameter and get the equation. Both will come out to be the same. By the way, if this is x equal to two and this is x equal to three, the x coordinate here will be two, x coordinate here also will be two. The x coordinate here will be three, x coordinate here will be three again. Okay. And this is y equal to one. So this is one, this is one. And this is y equal to two. So this is two, this is two. Okay. So which one you want to choose as your diameter? It is absolutely your call. So let's choose AC as the diameter and get our job done. So x minus x one, x minus x one, x minus x two plus y minus y one, y minus y two equal to zero. Okay. So this gives you the answer. Let's check. What does it give us? x square plus y square minus five x minus three y. And if I'm not mistaken, the constant term will be coming out to be eight. Correct. Now, as a matter of check, we will also find out the equation by using BD as a diameter. So in this case, I use AC as my diameter. Okay. Let's for a change now use BD as the diameter. Let's see whether are we getting a different equation. We hope not. We should not get a different equation. We'll get the same equation, but let's check. Let's check just for confirmation part. So with BD as a diameter, your equation will be x minus two x minus three. I don't see any change. Here also we had x minus two x minus three. Here also we are getting x minus two x minus three. So no change. And your y part will be y minus two y minus one. Oh, again, no change. This is also the same. Okay. So in expansion also, it should not give you a different answer. So we'll get the same result anyhow. Is it fine? Any questions? Any questions? Any concerns? Anybody? Okay. So we are going to do one thing. We are going to complete all the equations, all the type of equations. And then we're going to take more and more problems. So we'll have complicated situations also. So we're going to take more and more problems. So let's take the next type, which is called the parametric form of the equation of a circle, parametric form of the equation of a circle. Now before we move on to parametric form of the equation of a circle, it is very important for us to understand what is a parametric form and why do we at all need it? Okay. See, whatever form you studied, whatever equations you studied, whether it was the center radius form, whether it was the general form, whether it was the diametrical form, you realize that in those forms, there was a direct relationship written between your two variables, x and y, isn't it? Okay. So one single relation was written, which basically has to be followed by every point lying on that curve. Okay. So those form of the equations will be called Cartesian form. Okay. Those form of the equation will be called as the Cartesian forms. But however, if you write an equation in a way where you independently link x and y to a parameter, then it would be called as a parametric form. Let me give you an example. Let me give you an example. Let's say I take a standard form of the equation of a circle. Okay. Okay. As an example, let's say this is our standard form of the equation of a circle. So this is a circle whose center is at 00 or origin and radius is r. This is a Cartesian form. This is a Cartesian form. Okay. So now what I'm going to do is I'm going to write x as r cos theta and I'm going to write y as r sin theta. Now I claim or I basically suggest that this is a form by which you will end up getting the same equation if you eliminate theta. That means here I have written x and y in terms of a parameter. Now what is a parameter? Parameter is nothing but it's a changing constant. Many people say sir, it sounds like an oxymoron, changing constant. It is changing also and constant also. Yes, y is a constant. It is not a variable. Variable is x and y but it is a constant which keeps changing in the sense that you can change your theta and get different x and different y but for a particular theta, you have a particular y. So you may call theta to be like the Aadhaar card number of that point. See what is an Aadhaar card? Aadhaar card is basically our identification number. Is it constant for you? Yes. Jatin, do you have multiple Aadhaar cards? I don't think so. Unless until somebody is doing some kind of fraud and something, he will not be having multiple Aadhaar cards. Arya, do you have multiple Aadhaar cards? No, right? Only single Aadhaar card is there. So that number is a constant for you but does it mean that the same number will be given to everybody else? No, everybody has different numbers but for that person, it is fixed. Same way, try to understand the parameter also. So parameter is different for different points but for that point, it is fixed for x and y. Are you getting my point? Okay. So for example, on this circle, if you have a point, let's say r comma r, let's say no, r comma r cannot lie, r by root 2 comma r by root 2. Then for r by root 2 comma r by root 2, the theta value will be 45 degrees or pi by 4. So for that particular point, r by root 2 comma r by root 2, let me write it down here. So for r by root 2 comma r by root 2, your theta is going to be a fixed value. But if I change the point, let's say if I take a different point, let's say r comma 0, then my theta here would be pi by 2. Correct. So these points are like different, different people living on that particular curve. So think like, think like that curve is like a country and on that country, a lot of points are staying. So every point is known by its parameter value. Okay. That is the Aadhaar card number for that point. Is it clear what I'm trying to say? So instead of directly relating x and y to each other, you are relating them through a parameter. So this parameter connects these two x and y. Now many people ask me, sir, if a parametric form is given to us, will it be a sacrosanct parametric form? That means for this curve, is this the only parametric form? No. Parametric forms are changing. I mean, it is one of the suggested parametric forms. It is not a sacrosanct. That means all this, all the circle of this type need not have this parametric form. Then of course, the natural question will arise in your mind, sir, for the same circle, can you suggest another parametric form? Yes, why not? I can suggest one more parametric form for you. So this is another parametric form. Let me write it down. x equal to r by root 2 cos theta plus r by root 2 sin theta and y is equal to r by root 2, let's say cos theta minus r by root 2 sin theta. Okay. Now, in this, if you eliminate your theta or in other words, if you square them and add them, let's do that. Let's do that. Let's check. So if you do x square and y square, what you will end up getting? Let's check. You'll get r square cos square theta by 2, r square sin square theta by 2. Okay. Now, you'll also get two times this into this, but that will get cancelled with the term down over there. So I will not write it down. And this term will also give you r square by 2 cos theta plus r square by 2 sin theta. In other words, you'll end up getting r square cos square theta plus r square sin square theta, which means you end up getting r square only. Right. So when I eliminated theta, I ended up getting the same standard form, isn't it? So this could be another parametric form, another parametric form. So parametric forms are not sacrosanct. Please get this clear. One can write several parametric forms for the same curve. Are you getting my point? Okay. Now, why is parametric form used? Okay. So you will be thinking, sir, why unnecessarily we are learning parametric form? I am happy with Cartesian form. Why do we need a parametric form? See, parametric form has a purpose. And the purpose is to facilitate choosing a point on that curve. Okay. Let me give you a simple example. Where should I write? Where should I write? Where should I write? Okay. See, let's say I have a line. Okay. Let's say straight line. I'll take a very simple situation. Straight line. The straight line equation is x plus y is equal to 5. Okay. Let's say. Okay. Now, let me give you a small task. Choose a point on this line. How would you choose it as? I would like to know your response on the chat box. Choose a point on this line. Choose a coordinate on this line. How will you choose it as? No, a generic point. Choose a generic point on this line. Now, many people will say, okay, let the point be x1, y1. Okay. And unnecessarily they will say where x1, y1 should add up to give you 5. Now, see, what is the problem with this? The problem with this is unnecessarily you are introducing two unknowns, x1 and y1. And not only that, you have to always carry the burden on your head that x1 and y1 will satisfy this equation. Yes or no? Yes or no? So in coordinate geometry, lesser the number of unknowns, more simpler is your life or more free is your life. Okay. So here, if I have to choose a point, I would not choose it as x1, y1. I would not choose it as alpha beta. I would rather choose it as something like this. See, t comma 5 minus t. Where what is t? t is a parameter. In other words, for this line, I have used a parametric representation like this. Okay. Where t is a parameter. So if you automatically eliminate the parameter, you end up getting this Cartesian form. This is the Cartesian form. Okay. And this is your parametric form. Now, is this the only parametric form? No, I don't mean to say that. Please don't get me wrong. Now, don't be like, sir, for this, this is the only parametric form? No, this is one of the parametric forms. Right. I could also choose it like this. t plus 1 comma maybe 4 minus t. Well, this is also a work, isn't it? Right. So x equal to t plus 1, y equal to 4 minus t. This could be another way of choosing a point. Okay. Is it fine? Are you getting this fact that y parametric form is required? Now, if you have chosen a point like this, there is only one unknown t. So number of unknowns is only one as compared to two when I chose it as x1, y1. That is one benefit. Second benefit is I don't have to worry about x1 plus y1 being equal to 5 because for any t, x plus y will be equal to 5. That will always be satisfied. So choosing a point by taking a reference from the parametric form simple makes your life very, very simple while solving complicated questions. Are you getting my point? Okay. And hence it is also recommended that your parametric form should look very simple. For example, in our case, out of these two, will you prefer this parametric form or will you prefer this parametric form? Both are right. Both are parametric forms. Which one will you prefer? The first one or the second one? I would prefer the first one. Exactly. It looks so simple. So cute also. Who will write a point like this? Okay. So normally when we ask students to give a parametric form, we always recommend the students to give a simplest kind of a parametric form because the purpose of the parametric form is to choose a point. You never like to choose a point with ugly looking x and y coordinates. Would you? I don't think so. Correct. Is this clear what is a parametric form and why it is used? Later on, when we solve the advanced type of versions, you would realize that parametric forms work miraculously in solving many of the complicated problems. However, unfortunately, this is, you can say, unfortunate from CVSC part that we don't introduce this in our school curriculum. I don't think so. You would have done parametric form of the circles in your school. But it's a very important tool. So I always have this complaint from CVSC is that you don't allow the students to use shifting of origin and rotation of axes to solve problems. But that's a very important tool. You don't allow them to use parametric forms to solve questions. But that is also a very important tool. Okay. Anyways, for JEE and other competitive exams, we will be learning parametric forms. Meanwhile, any questions? Now, if you have understood this, can I ask you to suggest a parametric form for some curves? Will you be able to write a parametric form for this curve? Okay, let's do some simple exercises. Let me begin with the general form of a circle, the center radius form of a circle, my bad. So if this is your Cartesian form, if this is the Cartesian form, can you suggest, again, I'm using the word suggest because the answer may differ from person to person. Can you suggest a parametric form for this curve? That means, that means, can you write X equal to something in terms of a parameter Y equal to something in terms of a parameter? Can you do that? If yes, please go ahead and do it. And if you want, you can put your answer on the chat box as well. Very good, Jatin. So Jatin is suggesting something like this, alpha plus r cos theta and beta plus r sin theta. Wonderful. See, if you want to check whether this parametric form is right or not, just try eliminating theta. If you try eliminating theta from these two equations, it should automatically give you the Cartesian form. Isn't it? Let's try to check whether it is giving you the right answer. So if you replace or if you put your alpha to the other side, you get this. You put beta to the other side, you get this. Okay. And then square them and add them. Okay, so you end up getting r square, cos square plus r square sin square, which is as good as r square into one. And there you go. This matches with our Cartesian form, which I gave you. Yes or no? Okay. Now, many people ask me, sir, how did it get so fast? I mean, or how did Jatin get so fast? Jatin probably would be the right person to tell this, but mostly what people do, they treat this as capital X plus capital Y square equal to r square. And just now taking a clue from this, let me go to the previous board. Yeah. So from this board, we came to know your variable X is r cos theta, variable Y is r sin theta. So I can use that same strategy here also. So I can call capital X as r cos theta and capital Y as r sin theta. And now just replace your capital X with small x minus alpha. Okay. And capital Y with small y minus beta. And there you go, you can get, you get the same equation as what I wrote here. Is it fine? Okay, very good. So using this, we can also write down the parametric form or a suggested parametric form for the general form of the equation. So this is the general form. You can write down the parametric form for this like this X equal to, now see alpha. Alpha here is minus g plus r cos theta, this is r. Okay. And Y is equal to beta plus r sin theta, beta is minus f. And this is your r. Yes or no? So this is a suggested parametric form. Again, I'm not claiming this is the only parametric form. No. Like nobody can claim that this is the only parametric form existing. No, this is a suggested parametric form. Okay. However, these are, you know, very rarely been used. We mostly use the parametric form when there is a standard form of the equation involved. Okay. Is this fine? Any questions? Any questions, any concerns? Now, can we take some questions on parametric forms? Okay. Let's go to the questions based on this. Let's start with, let's start with this question. If the parametric form of a circle, okay. Now, actually the word the here is misleading. If I were the question center, I would not use the word the, I would use a, if a parametric form of a circle is given by, let's say the first one. Okay. Then find its Cartesian form, find its Cartesian form. Everybody try the first one and do let me know once you're done. Done for the first one. Okay. So whenever somebody asks you for the Cartesian form from the parametric form, okay, all you need to do is just one simple step, eliminate the parameter involved. Just eliminate theta from x and y. Your job is done. So the simplest way to eliminate theta is bring the minus four to the left and bring the minus three to the left in this case. Just square them and add them. Just square them and add them. When you square them and add them, you'll end up getting, we'll end up getting this. Okay. Of course, if you want to write them in the general form, you're most welcome to do that. So that'll give you x square plus y square plus 8x plus 6y equal to zero. Okay. So this is the answer to the first part. Now I would request you to do the second part as well. B. Okay. And just say you're done if you're done. Did I actually forget that? Jatin, see carefully. Right. They would get cancelled. No. Yeah. Okay. Satyam is done. So alpha is acting like the parameter in this case. So all you need to do is just try to get rid of alpha. I'm sure you must have done ample amount of such questions in your class 10 signometry if I'm not mistaken. Okay. Chintya is done. Neel is done. Very good. Okay. So you'll just do one thing, just square them and add them. So if you square them and add them, unlike, sorry, just like one of the questions that we did earlier, this will give you a square, cos square alpha, b square, sine square alpha. And you would also get, okay, let me write it here just to be more specific. You'll also get 2ab sine alpha cos alpha. And you'll also get a square sine square alpha b square cos square alpha minus 2ab sine alpha cos alpha. Okay. So on adding these two will anyways go for a toss. These two collectively will give you a square and these two will collectively give you a b square. Am I right? In short, your expression will give you something like this. Okay. Now, you must be thinking, sir, I mean, I feel this is a bad way to write the parametric form. Okay. See, you cannot say bad or good to it. But yes, it's a complicated way to write a parametric form. I mean, a normal person would not go with this parametric form. Rather, I could just write it as r cos theta or r cos alpha and r sine alpha. This would have been a better way to write it. Okay. So just wanted to tell you that this is another parametric form that could be written for the same curve. Okay. So this is another parametric form. Okay. So parametric forms are not fixed just like Cartesian form. Parametric forms can change as per the whims and fancies of the person who wants to write it. See, for the same circle, there are two parametric forms. This is one and the one given to you in the question. Two parametric forms are possible. Similarly, somebody could write more forms also. I can switch the position of sine and cos. It doesn't matter also. So that's the third parametric form. Are you getting the agenda behind this? Okay. Shall we try another one? Any questions? Any concerns? By the way, later on you learn that this is the equation of the director's circle of an ellipse. Okay. Anyways, we'll not talk about it because those are some concepts which we are going to study a little later on. So let's take another question. Meanwhile, anything that you would like to copy, write down, ask, please feel free to do that. Okay. All good? All right. Let's take another one. Take this one. Yeah. Find the parametric form for this circle. Again, this find is not the right word. Okay. If I have any questions, I would use the word suggest. Suggest a parametric form. Okay. So parametric forms are not fixed. Let's do this and give me a response on the chat box. If you feel it is too lengthy for you to type, you can just say I'm done. That is also fine with me. I mean, if I were to solve this question, I would actually first write it as a center radius form, which happens to be this, right? X equal to minus p and y equal to minus p. Where is the parameter in this case? Because p is not a variable. p is a fixed value that they have given to us. Harshita, you're getting confused. p here is a fixed value. It's not a parameter. They're asking you to write a parametric form for this. Okay. Okay. Let's see how complicated it is as you're claiming it's going to be a complicated form. I don't think so. It should be that complicated. So just try to compare this with X minus alpha whole square, Y minus beta whole square is equal to R square. For such cases, our parametric form normally happens to be or normally is taken as or it's a suggested parametric form. Something like this we take, isn't it? This is what we take for generally for this particular Cartesian form, isn't it? So here your alpha is negative p by two. So please just compare alpha is negative p by two. Beta is also negative p by two and radius is p by root two. Correct? No. So I would write a parametric form for this curve like this. X is equal to minus p by two plus p by root two cos theta and Y is equal to minus p by two plus p by root two sine theta. Is this fine? How did you get p by two? How did you get p by two? How did you get p by two? Okay. Completing the square, let me teach you completing the square. Okay. Now you understood. Good. Okay. So completing the square. So I took this together. I took this together and I started completing the square. I took this together. I started completing the square. Okay. Okay. Anything else that anybody is asking? I think Satyam, yes, more or less we have got the same thing. Satyam, is it fine? Any questions? All right. Now what are the different varieties of questions that we get apart from these equations? So many times the question that you would be, many times the question that you would be getting is different situations will be provided to you. For example, one situation that would be provided to you is like this question. Okay. The question has given to you two points and the line which contains the center. Okay. And then the question setter is asking you for the equation of the circle. So we are going to do a mixed variety of type of questions that is normally asked. All of them are a school level. Okay. Don't worry. They're not J level as of now. So let's do this question. Find the equation of a circle which passes through 4 comma 1 and 6 comma 5 and has its center on this line. Just say done if you are done or if you want, if you're so much interested in typing out your answer, you can do that. No constraints from my side. Harjit, do you mind sharing your answer? Okay. Okay. Very good. Very good, Neil. Okay. So there's so many ways to actually solve this question. I would like to discuss with you certain ways that students take to solve this question. See, in this question, what has been given is that your center is lying on some line. Okay. Let's make it. Okay. And the circle is passing through two points. Okay. Let's say I make one point here, 4 comma 1 and one point here, 6 comma 5. Let's call them as A and B. Now, in order to get the equation of a circle, you require its center and radius, whether directly or indirectly, you have to find that out. Isn't it? So many people, what they do is, or Satyam is getting slightly different answer. Anyways, we'll discuss it. What people do is, is they normally follow two types or three types of method. One method is they'll assume that there is a center on this line. For example, if I have to assume a point on 4x plus y equal to 16, I may assume it to be something like this. T comma 16 minus 40. Can I do that? Again, I'm using my parametric form. Correct? Isn't it? So if you put Xs, TY automatically becomes 16 minus 40. And then they would use the fact that C A and C B are equal because both are radius, and they will end up getting the T value. Once they get T value, they know the center. One center is known. You can find the radius by finding its distance from any one of the two points. Okay. That's one approach. This is approach number one. Okay. I'll write it down below this. Find T by comparing, comparing that's by equating C A with C B. Okay. So once you find that your center is known. Okay. Center is known. Find the radius. Okay. Which is your C A or C B, whatever you want to call it. So center radius form can be written. That is one way to do it. Another way to do it is, and see, you are most welcome to use any one of the forms. No method is right or wrong. Okay. You can use any one of the techniques to get your answer. Okay. Another approach which people use is, let me show that approach to you. Let's say this is your A. This is your B. This is four comma one. This is six comma five. Now, many people, what they do, they write down the equation of the perpendicular bisector to this line. Okay. So how do you write the equation of a perpendicular bisector? Very simple. You know the midpoint. You know the slope of A B. So hence, you know the slope of the perpendicular. So wherever this line, let's say I call this line as L1 and L2. So L1 and L2 solve them simultaneously and they will give you the center of the circle. Once you know the center of the circle, C A or C B can be used as the radius. Correct. Once you know the radius, once you know the radius, once you know the radius and you already know the center, you can write down the center radius form. That's another way to do the problems. There's another way to do it. The third way, which I'm going to use right now to solve this question. The third way is, let me write it as one, two. Okay. And the third way to do this problem is what I'm going to use right now to solve this problem. I will say, okay, let the circle be this, because I know this is the general form of a circle. Every circle should satisfy this particular equation. So all I need to do is find my G, F and C. That's all I need. That's all I need to finish this problem. So I can use the fact that this circle, first of all, passes through these points, 4 comma 1 and 6 comma 5. So 4 comma 1 will satisfy this equation. So when you put 4 and 1, if I'm not mistaken, I'll directly write down and give you 17 plus 8G plus 2F plus C equal to 0. And similarly, 6 comma 5, 6 comma 5 will give me 36 plus 25, 61 plus 12G plus 10F plus C equal to 0. Yes or no? So one equation is this, another equation is this. And one more equation I can say is that the center of the circle, which is minus G minus F, should lie on this line. Remember, for this circle, the center is minus G minus F and that should lie on that line. So can I say minus 4G minus F should be equal to 16. That is our third equation. Yes or no? Are you all convinced of these three equations? If no, please highlight. I will explain once again if you're not convinced. If yes, we can move forward. So now we are going to solve this simultaneously. For that, I'm going to do one thing. I'm going to subtract the first equation from the second. Let's subtract the first equation from the second. So when I do that, I'll get 44, correct me if I'm wrong, 44 plus 4G plus 8F equal to 0. Okay. In other words, I'll end up getting, let me just simplify this, 4G plus 8F is equal to minus 44. Okay. That means I will get G plus 2F. In fact, I don't need to do anything now. I will just add these two. I'll just add these two. So when I add these two, I'll automatically get 7F is equal to minus 28. So F is minus 4. Okay. F is minus 4. Now you put it in any one of them. How did I get the third equation? Third equation, you were not listening. I was saying minus G minus F is the center of the circle and that should satisfy this line. 4X plus Y equal to 16. So minus 4G minus 4F, sorry, minus 4G minus F is equal to 16. That's how I got it. Okay. Now put it in any one of them. Maybe you can put it in the third one. So you get minus 4G plus 4 is equal to 16. That means G is minus 3. Correct me if I'm wrong. Yes or no? So now these are the two of the three information which I needed. So I only know G. I only know G. I only know F. Now I want to just get a C. So for C, you can use the first equation. So you can say like this 17 plus 8G. 8G is minus 24 plus 2F. 2F is minus 8 plus C equal to 0. So C comes out to be 15. Correct me if I'm wrong. C comes out to be 15. So once you have got F, G and C, you can write down the equation of our required circle as X square plus Y square plus 2GX means you'll get minus 6X plus 2FY means minus 8Y plus C means plus 15 equal to 0. This is your desired answer. Anybody who got this? Anybody who got this? Let me check. Let me check. Harshita made a mistake. Satya made a mistake. Neil Matthew was the first one to get this right. Well done Neil. Absolutely right. Okay. So many people asked me, sir, do you recommend this method? See, there's nothing like recommending this method. But I find this easier because I don't have to find center radius and then convert it to center radius form and then convert it to general form because many times the options will be given to me in general form. So here I'm not messing with any kind of type of forms and directly getting my answer in the general form. So I just have to find my G, F and C. That's all I need. That's it. Nothing else. Is it fine? Any questions? Any questions? Any concerns? Let's take another type of question which is basically based on finding a circle passing through three non-colinear points. But before that, if anything you want to copy from here, no down from here, please do it. I'm giving you around one minute to copy things. See, I got the first equation by substituting 4 comma 1 in this because 4 comma 1, as you can see, it's a point lying on the circle, right Satyam? So if you put x is 4 and y is 1, what do you get? 4 square plus 1 square which is 16 plus 117 and you have 2, 4G which is 8G, 2, 1F which is 2F plus C equal to 0. That's how I got it. Similarly, the second one I got it by putting, putting 6 comma 5, yes or no? Okay, convinced. Okay, great. So let's move on to the next type of questions. This type of question basically involves you to find out the equation of a circle passing through, what is this? I chose some different equations. Yeah, find the equation of a circle passing through three non-colinear points. Now, since your class 10 days, you would have studied the fact that you can always pass a unique circle through three non-colinear points, isn't it? So now this is a question where the same type of concept has been now asked in a coordinate geometry setup. So through these three points, what is the equation of that unique circle that you can pass? Okay, please everybody give it a try and then we'll discuss the different methodologies that we adopt to solve this. Oh, very good, Satyam. Satyam is done. Okay, let's give Satyam a few more minutes to others. Okay, Satya, Neel, excellent, Arnav, Pai, very good. Done. Okay. So again, there are different methods to solve this question as I already discussed with you in the previous problem as well. One way that you could use is you could assume the center to be some h comma k or alpha comma beta, whatever you may call it. Use the fact that cr is equal to cp and cp is equal to cq. So use cr is equal to cp and cr is equal to cq. I think any two you can compare. So two equations, two unknowns, which is your alpha and beta, can you solve for it? Okay, once you know the center, you can find cp, cr, cq, all of them will correspond to the radius. So you know the center, you know the radius, hence the center radius form. Okay, that's method number one. Okay, method number two could be, once you have been provided with these three points, okay, let's say pqr, you may go for finding there equation of the bisector of the chords. So let's say this is one chord, this is another chord, and you find out the equation of the bisectors of these two chords. Okay, the meeting point of the perpendicular bisectors is going to give you the center. Okay, and once you know the center, you can find out cp or cr or cq, they will all be giving you the radius of the circle. That's another way to solve the problem. Okay, but the third approach is what normally we will be taking, where you will be assuming that let our circle be, I'm so sorry. Yeah, let our circle be x square plus y square plus 2gx plus 2f5 plus c equal to zero. That's our general form of the equation of a circle. And let's make these points satisfy these, these equations of the circle one by one. So if I put one comma one satisfying it, I will end up getting two plus 2g plus 2f plus c equal to zero, let's call it as first equation. Okay, then let's put two comma minus one, two comma minus one will give you five. Okay, plus 4g minus 2f plus c equal to zero, that's our second equation. Third equation is you can put your the point three comma two. Okay, so three comma two will give you a 13 plus 6g plus 4f plus c equal to zero. Now many people say sir, three equations isn't, won't it be a too much ask to solve? No, it's going to be very, very easy to solve it. Just take the difference of the two equations, one by one. Okay, let's let's subtract second minus first. Okay, just subtract second minus first. What do you get? What do you get? 2g minus 4f and I think when you subtract it, this is going to be a three equal to zero. Okay, so that's your number four equation. Then subtract three minus two. That's going to give you eight plus 2g plus 6f equal to zero. For it. Yes or no, that's your, that's your fifth equation. Oh, sorry. Are you? Yeah, that's a fifth equation. Now subtract these two, just do four minus five. Okay, so when you do four minus five, what do you end up getting is minus five minus 10f equal to zero. Yes or no, so f comes out to be minus half. Okay, a useful piece of information for us. Now put this in any one of them. Let's say I decide to put it over here. Minus 4f, minus 4f will be two. So g will be minus five by two. Okay, now let's put g and f in the first one. So I will end up getting two plus 2g. 2g is negative five. Okay, 2f is negative one. So can I say c will become a four? Correct, c will become a four. Any questions? So once I've got, once I've got g, f and c, my answer would look like this and check your answer as well. x square plus y square plus 2gx. 2gx will be minus five x. 2fy will be minus five plus four equal to zero. Is this fine? Any questions? Any concerns? Okay, now I will tell you a shortcut also since most of you are already aware of determinants, there is a yet another way to solve this question by using off determinants. Oh, but you don't know how to solve four by four determinant, right? Okay, but still I'll give you the formula. Maybe when we do four by four determinants in our class 12, then you can start applying the formula. Okay, so there's yet another way, fifth way to solve the question. Okay, sorry, fourth way to solve the question. So I'll just give you a formula. Equation of a circle, equation of a circle which passes through, which passes through three non-colonial points, x1, y1, x2, y2 and x3, y3 is given by, please note this down, it's a four by four determinant, x square plus y square xy1, x1 square y1 square x1 y1 1, x2 square plus y2 square x2 y2 1 and x3 square plus y3 square x3 y3 1 equal to zero. But the problem is it's a four by four determinant and you don't know yet how to solve four by four determinant. So we'll discuss this in class 12. Okay, in class 12, we'll talk about it. Okay, so this is the typical type of questions that you will be getting. I'll be taking a break right now. Okay, it's already six. So we'll just take a small break. Anybody who wants to copy this, please do so. I'll take a break. Okay, let's have a 15 minutes break. Right now it's 6.03. Let's meet at 6.18 pm sharp. Okay, we'll talk about more interesting aspects of this chapter. All right, see you on the other side of the break. Next concept that we are going to talk about is the concept of intercepts, intercepts, intercepts cut by a circle, cut by a circle on the coordinate axis, on the coordinate axis, coordinate axis. Okay. So let us say we have a general form of a circle and that general form of a circle cuts the coordinate axis like this. I'm just drawing a, let me use a different color maybe. Let's use a white color. So let's use a white color and make a circle. Okay, so let's say this is your general form of a circle, x square plus y square plus 2gx plus 2f5 plus c equal to zero. Okay, and I want to find out what is the length of the x intercept. This is called the x intercept. Let me call it as ab. Okay, so ab is called the x intercept. Just like we used to have intercepts in case of straight lines. Okay, so in straight lines, we basically used to find out from origin, right, origin to the point where it is cutting the x axis that used to be called as the x intercept. Right, in case of a circle, it is the part of the x axis which is trapped between the circles or between, in between the circle. Correct. Similarly, this length, cd is called the y intercept. cd is called as the y intercept. Okay, so now let us figure out what is the length of these intercepts in terms of gf and c. Okay, so this is given to us. Okay, this is given to us. This is given to us. Okay, and we need to find out what is what is the x intercept and the y intercept. Now, there are two approaches here that we can use. The first approach is the use of geometry. I'm sure most of you, I mean, are well versed with your geometrical aspect of a circle. So let's say this is my center. If I drop a perpendicular from the center onto this, let's say chord AB. Right, can I say AM will be equal to MB that means this chord is bisected by the perpendicular drop from the center. Let's say I call the center as P. Yes or no? Everybody agrees with that? Any questions? Okay. Now, let me ask you a simple question. By the way, many people are not able to answer the simple question. Tell me what is this length PM? Who will tell me what is the distance of P from the x axis? What is this distance PM? Write it down on your chat box. What is the length PM? Okay, Arya, what if minus F was a negative quantity? Sorry, some of you have said F, some of you have said minus F. Yes, the right way to answer is what Satyam has said, it's mod of F. Okay, please note any kind of a distance. Don't use the coordinates as it is because coordinates may be negative. What if my center of the circle is negative quantity? Right. I mean, both x and y coordinates are negative. Right. It's not a good idea to write it as F or a minus F, better to write it as a mod F. Okay, good. What about PA length? You'll say sir PA length is the radius, which happens to be under root G square plus F square minus C. Yes or no? Now, so can I, can I use Pythagoras theorem on this triangle PMA? So by Pythagoras theorem, by Pythagoras theorem, can I say PA square is equal to PM square plus AM square? Yes or no? PA square is the radius of the circle square, which is G square plus F square minus C. PM square will be just F square, just now you told me. And AM square, I don't know, I have to still find out AM square. Right. So let's cancel out F square, F square. So from here I can say AM is under root of G square minus C. Is it correct? Yes or no? Now I know that M is the midpoint of AB. So AB is twice of AM, which is nothing but twice of under root G square minus C. Yes or no? Okay. Please note this down. This result is your x-intercept. So please note this down. This is your x-intercept. Okay. This is your x-intercept. Any questions? So many, many people ask me, say, do we have to remember this result? If you do, you save your time. That's it. I mean, there's no hard and fast that you have to remember it. But if you do, you happen to save your time. If at all it is required in some question. Okay. Now I would request you to figure out the y-intercept in a similar way. Let me see who gives me the y-intercept. So I'm writing similarly, y-intercept is equal to what? I'm basically relying on you to give me the answer. Please write down on the chat box. What do you think will be the y-intercept in this case? Well done, Satya. Well done. That's absolutely right. Absolutely right, Harshita. So it's two under root f square minus c. Well done. Okay. So both these results have to be noted down. Please note it because it will be useful in solving a few questions wherever it is needed. Now remember when I was deriving this result, I told you this is one of the ways to do it. Right. So this is the geometrical way of solving the problem. So what is the another way to do it? Okay, let's look into another way of doing it. And this another way is going to be helpful when we do chapters like parabola ellipse, et cetera. Okay. So another way to do it, let me write it as all it up. Okay. All it means another way to do the same task. So now all of you please pay attention. Okay. Let's say this is my, this is my x-axis and this is my y-axis. Okay. Now x-axis, x-axis equation is what? x-axis equation is y equal to zero. Right. And let's say this circle is x square plus y square plus two gx plus two f y plus c equal to zero. Okay. Now just put y equal to zero in this expression. It gives you x square plus two gx plus c equal to zero. That means what you are doing, you are simultaneously solving these two equations one and two. Right. So in order to know the x intercept, what you are going to do in this case is you are going to put your y as zero in the equation of the circle. So when you put the y equal to zero in the equation of a circle, you end up getting this expression. Okay. Now we know that it's a quadratic in x. Okay. It's a quadratic in x. It's a quadratic expression in x. And this will have two roots. This will have two roots and let's say the roots are x1 and x2. Okay. So x1 may be the x coordinate. You can choose this to be your x1 comma zero and x2 is the x coordinate of this point. Okay. So please try to understand here. I'm trying to claim that x1 and x2 are the roots of this quadratic expression. Okay. Just give me a second. Okay. So basically what I'm trying to say is that this x axis, this x axis cuts the circle at two points A and B. And I'm assuming that A and B, the x axis, the abscissa coordinates are x1 and x2, which will be actually the root of this equation. Correct. Now, can I say the length of AB is modulus of x1 minus x2? Am I right? Does everybody agree to me on this that the length of x1 and x2, so the length of AB will be modulus of x1 minus x2? Agreed. Okay. So can I not write this as under root of x1 minus x2 square? Correct. Agreed. Can I not write this as x1 plus x2 square minus 4x1x2? Correct. Can I not write it as, now see everybody in a quadratic, what is the sum of the roots? Minus B by A, isn't it? So in this case, it will be minus 2g by 1, which is actually minus 2g. So square of that will be 4g square. Okay. Minus 4x1x2. Now, product of the root is c by A. Now, in this case, c is c and A is 1. So it will be c only here. Just take a 4 outside, it will become 2 times under root g square minus c. The very same result as what we got by the use of geometry. Okay. Now, many people say, do we need to know this approach because I'm fine with geometry. Can I use geometry every time? See, of course, you can use it for a circle. But every time geometry is not going to save you, especially when the figures become un-symmetrical in nature. Okay. Or the figures become, let's say, if it is a circle, you can manage it. But what if it becomes a parabola? What if it becomes an ellipse or something like that? In those cases, you would realize that this approach will be more handy. Are you getting my point? So please also appreciate this method of finding the intercepts. Okay. Note it down and let me know if you have any doubt anywhere. Do let me know. Similarly, you can also do the same for finding the y-intercept. In y-intercept, you have to put x as zero because you're trying to find out where it is cutting the y-axis. So on y-axis, x is zero. So when you put x as zero, you end up getting a quadratic in y. So you know your quadratic equation skills to get it. Okay. So please make a note of this and let me know if you are done with copying this. Yeah. Yeah. Okay. I have a small question for you all just to make you think a little bit. Okay. Let's say my circle is like this. Okay. My circle just happens to touch both the x-axis and the y-axis. Let me make it in white. I've already used yellow color. Just give me a second. Yeah. So let's say my circle touches both x-axis and y-axis. Okay. In this case, what is your x-intercept and what is your y-intercept? Leave your response in the chat box. What is the x-intercept in this case and what is the y-intercept in this case? Quickly, quickly, quickly. I would request everybody to give their response. If it is just touching the x-axis and just touching the y-axis, what is the x-intercept and what is the y-intercept? Vishal, you're telling me the center of the circle. How is that the intercept? Intercept is that part which is trapped within the circle from the x-axis. No, no, no, Vishal, you're not understanding my dear. See, this trapped part is what we call as intercept. This is x-intercept. This is y-intercept. What is the trapped part of the x-axis in this situation when your circle is just touching the x-axis? Zero. Zero. This is also zero. Are you getting a point? When your circle just touches the x-axis, x-intercept is zero. My intercept is zero. Intercept means the part of that x-axis which is trapped within the circle. That is called intercept. What you are telling me is what is this distance? No, this is not the intercept. That is true for a case of a straight line that doesn't work in case of a circle. Let me tell you this. Getting my point. Okay. I'll give you one more question. Think and answer. Think and answer. If the circle is such that it is touching the x-axis, let's say it is neither touching the x-axis nor the y-axis. In this case, what is the x-intercept and what is the y-intercept? All of you are actually stumped. Both will be imaginary. Let's see why it is imaginary. I'll tell you a simple situation. If your line is cut by the circle, whether x-axis or y-axis, let me just take a generic case. Let's say this is a line. If it is cut by the x-axis, we say that this intercept is positive. In this case, your intercept is positive. If your line is just touched upon by the circle, or let's say the line just touches the circle, let's make it in yellow. Be consistent with the color. If your line just touches the circle, your intercept is equal to zero. But if your line is not even touching that line, that means if it is something like this, let's say it is hanging up like this. In such case, your intercept is imaginary. And you can calculate that it will come out to be imaginary. Let me take a scenario. Let's take an example of the situation. Let's say this is a circle whose center is at 2.3 and let's say the radius is 1. We are all convinced that it is not going to touch either of the axes. If the center is at 2.3 and radius is 1, aren't we all convinced that it is not going to touch any of the coordinate axes? Let's do a simple exercise here. Let's get the equation of the circle first. Let's get the equation of the circle first. Let me just simplify and write it. 4 plus 9, 13 plus 12 equal to 0. Now, please find out under root 2 g square minus c in this case. So you will get 2. Now g here is minus 2. So square of that will be 4 minus 12. What do you see? 2 under root minus 8. What kind of a quantity is this? Is this a real number? No, I don't think so. It is an imaginary number. Yes or no? Try finding 2 f square minus c. What do you get? 2 under root. f square is 9 minus c. See what do you get? 2 negative root 3 under root of negative root. Again, imaginary. So this is the wrong conception that many students carry is that if the line is touching, then also the intercept is 0. And if it is not touching, then also the intercept is 0, like most of you answered, but no. That is not the case. The case is if it touches, intercept is 0. But if it doesn't touch, forget about cutting, it even doesn't touch, then your intercept will be imaginary. Get this point right. Is it clear? Any questions? Any questions, any concerns? Okay. Should we take few questions? Okay. Let's take a very simple formula-based question. Find the equation of a circle whose diameter is the line joining these two. Even if you don't give me the first part, I'm fine. What I'm interested in is in the second part. Find the intercept made by it on the y-axis. Anybody with the response of root negative 4a? Okay. I mean, it may happen that you get a negative intercept, but I'm just doubting that it will happen in this case. Okay. Okay. Thank you, Jatin. Thank you, Satya, for giving the entire answer. Okay. Let's discuss this out. Not a rocket science. Find the equation of a circle whose diameter is the line joining these two points. So we have already seen the equation of the diameter, sorry, the equation of the circle would be this. Am I right? Any questions with respect to that? On simplification, this gives us x square plus y square minus 8x minus 2y minus 48 minus 48 minus 3, which is minus 51 equal to 0. Right? This is the equation of the circle. Now, what is y-intercept? y-intercept formula, we have already discussed it. It's 2 times under root f square minus c. So it is 2 times under root. What is f square? f square is 1 minus minus 51. That makes it 2 under root of 52. Under root 52 is 4 root 13, if I'm not mistaken. Okay. So this many units is going to be your answer. Is it fine? Well done, Jatin. Very good. Neel also. Absolutely right. Let's take another question. Find the equation of a circle which touches the axis of y. Oh, that's a nice thing of saying y-axis at a distance of four units from the origin and cuts the intercept of six units from the axis of x. Okay. I have drawn some diagram on the screen right now. See, the question says it just cuts the intercept of six on the x-axis. Okay. And it touches the y-axis at four units from the origin. Now, if you see the case, I have four cases for it. First case, I can write this as your second case. Okay. Third case and fourth case. In each of them, you'll see that you could have a possibility where this is six and this length is four. Correct. So four answers are supposed to come out for this question. However, I will be happy even if you get the first one. Okay. Because second, third, fourth one can be similarly be found out to answer your question. Jatin, yes, you're right. The y-intercept in this case could be taken as zero because it is just touching the y-axis. Yes. Anybody? Okay, Satya. Let's give a couple of minutes more for others who are trying hard. So just give me the equation of the first circle. I don't need the others. Others, you can similarly find it out. Don't worry about that. Yeah. Okay. See, the first approach that I would like to take is a very generic approach. I will assume that I don't know any readymade formula. Okay. Some of you are asking what is this intercept? Let's say I just use my generic, you know, basic funda to solve this question. So if you know that this is four, can I say, can I say the y-coordinate of this point will be four? So it'll be something like alpha comma four. Do you all agree with me on that or not? Correct. Right now. Now, you already know that the x-intercept is six. So this is three and this is four. So what is radius then? Can I say the radius will become five units? Do you all agree with me on this as well? Correct. So once you know the radius, okay, now can I say this length is also five units? So alpha is also five. Correct. In that case, you know the center of this circle is five comma four and you know the radius is five. So what is stopping us from writing the center radius form? So x-5 the whole square, y-4 the whole square is equal to five square. Simplify this x square plus y square minus 10x minus 8y plus 16 equal to zero. That's the answer. Now, can I use this to get the answer to the other parts also quickly? Yes, why not? All you need to do is just change these signs. So you keep up plus minus plus minus here. So you have four combinations. So one with a minus minus, one with a minus plus then with a plus minus and then with a plus plus. So all four answers will come out from here. Yes, I know. Any questions? Any questions? Any questions? Okay. Sorry, let me go back to the previous question. I would like to do this question by a formula also. Many people will be thinking that, sir, how can we solve this by using our formulae? Okay. See, first of all, the question center has provided me with the fact that its x intercept is six. So this is six. And y intercept would be zero. Okay. As per our question. Yes or no? Is it fine? Any questions? Any concerns with this? So this leads to two situations. g square minus c under root is three. That means g square minus c is nine. And from here, f square minus c is equal to zero. Yes or no? Yes or no? Okay. So now two equations I have already got. Two equations I have already got. How will I get the third condition? How will I get the third condition? How will I get the third condition? The fact that you know that this is four, you already know that minus f modulus is four. That means f modulus is four. That means f is plus minus four. Correct. If you know your f is plus minus four, from this equation, I can say 16 minus c is zero. So c is 16. So f is plus minus four, c is plus minus 16. And from here, the first equation, you can get g square is equal to, or g square minus 16 is equal to nine. So g square is 25. So g is plus minus five. So you know your g, you know your f, you know your c. So the equation of a circle is very obvious from there. x square plus y square plus minus 10x plus minus four y, sorry, plus minus eight y plus 16 equal to zero, which basically matches with our given answer as well. Is it fine? This is when you want to use the formula. This is when you want to use the formula. Is it fine? Yes. I think you also have a very valid point. Jatin, he said the circle would satisfy zero comma four. That's also can be used. Okay. Can we now go to the next question if this is done? Okay, let's do this question. Find the equation of a circle which touches the axes and whose centers lie on this line. Yes, done. Anybody? Okay. So again, if I want to solve this question, if I just want to solve this question by our basic fundamentals, basic fundamentals. Okay. Now, as of now, I'm assuming that the set the circle is in our first quadrant, but just by looking at these figures, I cannot comment that. Okay. Now, what I know is that the circle touches both the axes and its center lies on this line. Now, let's say the center here is okay. If I want to solve this question from common sense point of view, can I say the coordinate of this particular point should be of the nature alpha comma alpha or it would be of the nature alpha comma minus alpha. Why? Because see minus G and minus F modulus should be same, which means G and F could be either same or they could be negatives of each other. Right. So if I assume this coordinate to be alpha, other would be either alpha or minus alpha. Yes or no. So either they would be of equal value or they would be of opposite science. Okay. Now, many people say why not plus alpha minus alpha plus alpha plus alpha minus alpha minus alpha see alpha itself will can have positive negative science. So don't worry about all those permutation combination. Just worry about the fact that they could either be exactly equal to each other or negative of sign whether both are positive, both are negative that is that will automatically be taken care of not to worry. Okay. Now, this alpha, let's say I assume alpha comma alpha is the center. So let this be the center. So this should satisfy x minus 2y equal to three. That means alpha is negative three. Correct. That means I could have a center as negative three comma negative three. Correct. And if this is the center, what will be the radius? What will be the radius? Radius will be mod of alpha, which is actually three. So if you know the center and if you know the radius, this becomes the equation. Correct. In other words, this is your equation. Am I right? This is one answer. So this is one of the possibilities. What about if the center is alpha minus alpha? Let's go. Let's take that also. So let the center be let the center be alpha minus alpha. So if it is alpha minus alpha, you can put it over here and it will become alpha plus two alpha equal to three. So alpha could be one. In other words, your center could be your center could be one comma minus one and your radius could be your radius could be mod of one, which is one Excel. So in such case, the equation is going to become x minus one, the whole square y plus one, the whole square is equal to one square, which is nothing but x square plus y square minus two x plus two y plus one equal to zero. Okay. So only two possibilities are there. Only two answers are possible. No third answer is possible. Can it be minus three comma three? Can it be minus three comma three? Does it satisfy this equation, Harshita? Does minus three comma three satisfy this equation? Definitely you yourself can answer this. Oh, you asked me before. Okay. Okay. Is this fine? Any questions? Now let us say somebody wants to solve this question just by using the formula of intercept. Then how do I solve this? As I told you, if the circle is touching both the x-axis and the y-axis, the x-intercept will be zero. That means this is equal to zero. y-intercept would be zero. That means this is also zero. And apart from that, minus g and minus f is going to satisfy the equation of this line because the center lies on this line. Isn't it? So minus g minus f is going to satisfy this equation. So you'll end up getting minus g plus two f. Okay. Minus g plus two f equal to three. Yes or no? So see, let's use these three equations to get our answer. So from the first one I get g square is equal to c. Second one I get f square is equal to c. Okay. And this basically implies that g square and f square are equal. That means g is equal to plus minus f. Okay. So let's say there are two cases coming up. g is equal to f. Second case, g is minus f. Okay. So if g is equal to f, then this will become minus g plus two g equal to three. That means g is equal to three. Correct? So three comma three is a center. Okay. Anyways, g is three. So f is also three. And c is going to be nine. Correct? If I take the second scenario where g is minus f, in that case minus g, and this is going to be minus two g is equal to three. So g is going to be minus one. So f is going to be one. So c is also going to be one. So one equation you can get from a, you know, using this, another equation you can get using this. Same answer you will get, not different. Isn't it? Okay. So I have, many people will ask me, say which approach you prefer. I will always prefer an approach which is coming from your basics because those are going to stay for you with you for long. Even after two years when you're actually writing these papers, right? The basic concepts are going to stay with you. Anything which is like over and above the basic concept can only be retained to practice. Okay. Okay. So you need to practice this out. Is it fine? Any questions? Any concerns here? All set? Okay. Now, we are not going to do a lot of things, but we are going to definitely wrap up this topic with a small concept which is called intersection of a line with the standard form of a circle. Intersection of a line, y equal to mx plus c with the standard form of a circle, with the standard form of a circle. Okay. Let's say we take this as our standard form. Okay. Just to tell you, I mean not to scare you, but just to make you more aware, circle has got almost, you know, four times more concept than what we have covered today. So basically this is just a three and a half hour session and whatever we covered, four times more is yet to be covered in circles. It is such a big chapter. Okay. If I just list down the topics that you're going to cover, you will be like, you know, there's almost 17, 18 topics that will come up in circles. Okay. But I will not be covering them in, you know, near future, I'll be covering them only after your semester exam is over because I want everybody to first be securing their marks in the school exams and then we'll, you know, take that up. Okay. So this is one of the concepts which is, you know, important. We'll talk about it. So in this concept, we are going to talk about how does a line interact with the circle? How does a line interact with the circle? Okay. A line can interact with the circle in three ways. Either it can cut the circle or it can touch the circle or it can be like going far from the circle. Okay. So our case is a very simple case where we have taken the circle to be this. Okay. And let's say this is our center of the circle. We already know this is our line y equal to mx per c. So this first situation is called the tangent situation or sorry, secant situation. I'm so sorry. The first situation is called the secant situation. So secant is a line which cuts the curve at two distinct points. We already know that. This situation is called the tangent situation. You already know tangents. Okay. So it touches the curve at two coincident points. And this third situation is called nsnt case. Can anybody guess the full form of nsnt? Anybody can guess what is nsnt? Very good answer. Nice guess. Neither secant nor tangent. Okay. So now I would request you to take this as a question. Okay. And get me what condition, under what condition the line will be a secant. What is the condition when this line will be a secant? Okay. If you want to tell me in words, so this is the situation which should be satisfied for the line to be a secant to the circle. What will that condition be? If you want to phrase that in words, of course, we'll convert that to mathematical language. But in English, if you have to phrase that condition, what will you say? You'll say, sir, the distance correct. Okay. Fine. So Satyam has a point intercept must be greater than zero. But do you have a formula to find the intercept of any line with the circle? Remember, your intercept was only for x axis and y axis. Another way to put in another way to express the same situation, Satyam, think about it very simple. I have made some diagram on the screen. You can use that diagram to tell me. Right. Simple. Distance should be less than the radius. Okay. So in this case, your radius is a. Okay. So distance should be less than the radius. And what is the distance? What is the distance of origin from this point? So zero minus m zero minus c mod by under root of one plus m square. There should be less than a correct. Which means mod c by under root of one plus m square is less than a which means mod c is less than a under root one plus m square. That means c square is less than a square one plus m square. Yes or no. So please note this down. This is the condition. This is the condition. Okay. Let me write it down. This is the condition for secant. Okay. Any questions here? What is D in this? D is the distance of the center from the line. Okay. I've already shown it in the diagram. I thought from the diagram you can read this. So now it is no surprise what would be the condition for tangent. D will be equal to a. Correct. And if you write it down, you will get exactly the same thing is just that this less than equal, less than sign will become equal to sign. So your condition for the line to be tangent is this. Okay. Please note this down. This is very important. Okay. Then we call this as condition for tangency. This is called condition for tangency. Please note this down. Okay. Third case is when the line is nsnt case. That means it is neither secant nor tangent case. So what will happen in this case? You will say it's a simple D will be greater than that is to say c square will be greater than a square one percent square. Yes or no. Okay. So out of these conditions that I've given to you, please understand that these conditions are only meant when your equation of the circle is a standard form of a circle. It only works for a standard form of a circle. It only works for standard form of a circle. Okay. In case you want to know the conditions, if your circle is the general form of a circle, you need to repeat the same thing with the general form of a circle. Are you getting my point? So one mistake which I've seen people committing here is that whenever a question comes, okay, no matter whatever is your circle, they start using these conditions. No, these conditions are very, very, you can say specific conditions. It only works when your circle is a standard form of a circle. It is not going to work if your circle is a generic form of a circle. Okay. For that, again, different set of conditions will come up. I mean, condition will be same, but expression for that condition will differ. Don't use the same condition for all the cases. That's a mistake which most of the students will commit in the beginning. Are you getting my point? All right. Let's take some questions based on the same. Let's take some questions. Let me begin with this question. For what value of lambda will this line be a tangent to this circle? For what value of lambda will this line be a tangent to this circle? Please give me a response on the chat box. Okay, Jatin, Satya and I think one more person has responded. Harshita. Okay, three people, three different answers. Okay, Satya. Okay, Jatin, no issues or to change your answer. It's fine. Okay. Let's discuss. So right now, if you realize looking at the question that this is the standard form of a circle and this is a y equal to mx plus c kind of a situation. So just compare, oh, I'm so sorry, mx plus c. I forgot to write a x. Yeah. So this is a situation where you have a standard form of a circle. Okay, and you have a line y equal to mx plus c. So where is the line and tangent to this circle? The condition was c square is equal to a square 1 plus m square. So remembering this result saves a bit of your time. Now c is your lambda. I don't know that. I need to find that out. Okay, a square is 5 and m is 2 square. So lambda square is 5 into 5, which is 25. So what is lambda? Lambda is plus minus 5. Okay. Now many people ask me, sir, why two values of lambda comes out? Why not only one comes out? See, you will get one value of lambda as 5 where the tangent will be like this, maybe something like this. Okay. And another value of lambda will come when your, sorry, another tangent will come when your lambda is negative 5 like this. How are you getting the point? Is it fine? Any questions? Any questions, any concerns? Okay, let's take some questions in general also. For example, find the point of intersection of this line and this circle. Yes, anybody? Okay. Okay, Satyam. Oh, the figures are that ugly. Second coordinates seems to be so ugly. Okay. Anyways, we'll check it out. We'll check it out. How about others? Anybody else who could get it? See, in plain and simple words, if I have to assess this situation, it's like solving these two equations simultaneously. Right? So you're solving this linear equation of a line with this second degree equation of a circle simultaneously. That's what you're doing here. Right? Okay. So all you need to do is, I mean, let me put my y in terms of x. So your y square will be what? 18 minus 2x by 3, the whole square. Okay. Let's try to simplify this. If I'm not mistaken, it will be 9x square. And this is going to be 4x minus 9, the whole square is equal to 25 into 9, which is 225. Correct? Yes or no? Let's try to simplify this. So if I'm not mistaken, it'll get 13x square. And you'll end up getting minus 72x. Okay. And 81 into 4, 81 into 4 will be 324. So 324 on this side comes out to be 99 if I'm not mistaken. Yes or no? Correct? Is this factorizable? I think this is factorizable as minus 39 and minus 33. Check it out. So take 13x common x minus 3, take minus 33 common x minus 3 equal to 0. Oh, yes. You're getting quite ugly figures over here. Okay. Yeah. So x value could be 3 and x value could be 33 by 13. Okay. Now, in order to keep the y coordinate, you just have to use this equation again, 2x plus 3y equal to 18. So in 2x plus 3y equal to 18, put x as 3, you get 6 plus 3y equal to 18. So y is equal to 4. So one point is 3 comma 4, absolutely right. Okay. And next point is 33 by 13. So if I'm not mistaken, it will be 3y is equal to 18 minus 66. Okay. How much is, how much this has come out to be? Let's check it out. 18 into 13, that will be 24, 1, 11, 4, 3, 2, 234. Okay, 234 minus 66. That's going to come out to be 168. Am I right? So y will come out to be 56 by 13. That means the next point will be 33 by 13 comma 56 by 30. Is this fine? Any questions? Any questions? Absolutely right, Satyam. Perfect. No mistake at all. Is this okay? Any questions? One last question we can take for the day. Okay. So when I was talking about the concept of condition for secant, somebody told me that the intercept left by the line in the circle would be a non-zero quantity. Correct. So in regards to that, I have a question for all of you. Find the length of the intercept of this line by this circle. That means there's a circle and there is a line. Okay. 4x minus 3y equal to 10 and x square plus y square minus 2x plus 4y minus 20 equal to zero. So what is this intercept is what we need to figure out. Absolutely Satyam. You're right. Should we discuss it now? Done. Okay. This center is 1 comma minus 2. Correct. And what is the radius of this particular circle? What is the radius? Radius is 1 square plus 2 square minus minus 20, which is plus 20, which is nothing but 5 units. Isn't it? The radius is 5 units. Now in order to find the intercept cut by this line or in order to find the intercept cut on the line by the circle, we need to find this distance out. Correct. And for that we need, let me call this as cmp. For that, we need cm. cm is the distance of the center from that given line. So let's find out the distance of the center from that given line. So it'll be mod of 4 into 1 minus 3 into minus 2 minus 10 mod by under root of 4 square plus 3 square, which is 5. Correct. In this case, you realize that this answer is actually a zero. Isn't it? If this is zero, which means this line is actually passing through the center of the circle itself. And in that case, what is the intercept? Intercept is twice of the radius, which is 10 units in this case. Yes or no? Yes or no? Is it fine? So if at all it was a case where the line was not passing through the center, then of course I would have found mp and I would have doubled it up. Mp, you can find it out by Pythagoras theorem. So you could use Pythagoras theorem to find mp and just double up your answer to get the length of that intercept. Is that fine? Any questions? Okay. So dear boys and girls, so we'll stop here. Okay. Next class when we meet, which happens to be on, which day? It happens to be on, on Friday. Yes, next Friday. Next Friday, no Tuesday will be chemistry again. Okay. So next Friday, I'll be doing with you mathematical reasoning. Okay, MR. And then the Saturday class. No, not for from now. The chemistry class will come back to Saturday once these three weeks are over. Okay, only applicable to these three weeks, not forever. Okay. So next Friday, we are meeting with mathematical reasoning and next Saturday, we'll take up parabola. Is that fine? So bye-bye. Signing off. Take care. Good night. Stay safe. Okay. Take good care of yourself. Thank you. Bye-bye.