 So last class, we ended our discourse, ended our discussion on chord of contact, right? So I was basically talking about pair of tangents, pair of tangents and chord of contact, and chord of contact. So I started my discussion with the chord of contact. So let me quickly make the diagram for the same. So chord of contact is what chord of contact is basically a chord which connects the point of contact of two tangents drawn from an external point onto the circle. So if I draw two tangents from this point onto the circle, okay? The line joining the point of contact, this line is basically called as the chord of contact. So by the way, the word chord here is a misnomer because it is actually a secant line to say, okay? It's not like it's only limited to these two points of contact A and B, no. It's a line, okay? It's a line, it's a secant line which passes, which cuts through the circle, okay? But we call it as a chord of contact because it basically carries within itself the chord which is obtained by connecting the points of contact of these two tangents drawn from an external point, okay? And we had also discussed that, we had also discussed that if the equation of this circle is X square plus Y square is equal to A square, then chord of contact, the chord of contact equation becomes X, X1, Y, Y1 equal to A square. Very much similar to that of the tangent equation but the only difference is it would have been a tangent had X1, Y1 were on the circle but X1, Y1 is not on the circle. Hence, the same equation would now be representative of the chord of contact, okay? And when you had a general form of a circle, let's say X square plus Y square plus 2GX plus 2FY plus C equal to zero, then the same chord of contact equation will now be XX1 plus YY1 plus GX plus X1 plus FY plus Y1 plus C equal to zero, by the way I was using R for the radius so I'll continue using it because should not make anybody confused, okay? Now if you look at the both scenarios, basically you realize that both these actually hint at the fact that the equation of the chord of contact, the equation of the chord of contact is given by T equal to zero. This is something which you need to keep in mind because if you know your circle, you know your S, when you know your point, you know your T, just equate to zero, you'll end up getting the equation of this chord of contact. So now we'll do a bit of analysis on this, a one common question that comes, in fact few common questions that come on with respect to chord of contact is number one. If I draw a circle with circumscribes, P, A and B, okay, so let me write it down. What is the equation of the circle? What is the equation of the circle? That circumscribes, that circumscribes are here. Circumscribes, P or you can say triangle P, A, B. Triangle P, A, B. How would you find out? Anybody has any idea about how to find out a circle with circumscribes? Again, let me make the figure. Once again, I'll just take a copy paste of this because I don't want to make this whole thing over and over again, let me just take this part, yeah. Very good, I'm getting some few leads how to find that out. Everybody, please contribute. How would you get the equation of a circle with circumscribes, the triangle P, A, B, A, B, P, A, B, with circumscribes, the triangle P, A, B. In other words, I want to know what is the equation of the circle? Let me use a white circle for this. I mean, let me try my best to fit in a circle here. There you go, I did a good job here. Yeah, yes, so I have got few hints. Ashutosh says P, A, B, O is a cyclic quadrilateral P, A, B, O, okay. O means the center of this circle, okay. Center of this circle, okay, let me do one thing. If I make the position of the circle things, position of the center things will be very obvious. Okay, so now most of you have actually cracked this. The center of this circle basically should be on this circumscribing circle. So the center of the red circle should actually lie on this white circle. Now why is it that, very simple. If you see, this is a tangent to that given red circle, right? So if I draw a line connecting the center of this red circle to this tangent, it must be perpendicular, okay. Now if it is perpendicular, then basically P, O must be the diameter. P, O must be the diameter of the circle, right? Do you all agree? And this was basically a type of construction that you use in your class 10th also in order to make two tangents from an external point to a circle, correct? So actually what did you do? You took the point, you took the center of the circle and you actually bisected it to get the center. You made a circle out of it and wherever the circle cut the red circle, you used to connect those from this external point. That is how we used to construct our tangents from an external point to a circle. I hope all of you recall that, class 10th days, okay? Geometry, yes, isn't it? Yes or no? How many of you remember that? Yes, some of you definitely remember it. So now if your center of the circle is known and this point is known, let's say a center of the circle is minus G minus F. I'm assuming that my circle is the general form of a circle, general form of a circle. Then the equation of this particular white circle would be nothing but the diametrical form where you know the diametrically opposite x. So what are the diametrical form of any circle? X minus X1, X minus X2. Now remember, X2 will become minus G. So X minus X2 will become X plus G plus Y minus Y1 times Y plus F equal to zero. So this becomes your equation of the circumscribing circle. Okay, so please keep this mind. In fact, there's no need to remember the formula. What you need to remember is what is the core principle behind it? How do we get the circles equation? So basically the circle center, the given circle center and the external point will behave as the diametrically opposite points. And once you know the diametrically opposite points, nobody can stop you from getting the equation of the circle. Is it fine? Another type of question that comes, why it has to pass? Okay, it has to pass because if I draw a, if I draw a line connecting the center, see my circle has to pass through A and P. That is for sure, right, Ravav? My circle has to pass from A and P because it has to circumvent, right, P or B. And you know that from the origin, if you draw a line to A, it will be perpendicular, right? If it is perpendicular means what is going to happen? On a circle, on a circumference of a circle, if you have two points, let's say P and A. And there is some point that you are basically drawing which makes 90 degree with this. So can that point lie here? Can that point lie here? Can this make 90 degree or can this make 90 degree? No, right? That point has to be lying on the circle. So only this line can make a 90 degree. And that can happen only when this is the diameter. Now getting the point, why I chose O to be a diametrically opposite point and hence lying on the circle. You imagine as if P and A are on the circle and there is a point from where you are drawing a 90 degree. And that point has to be on the circle. And it goes the same for the other direction as well. It goes for the other direction as well. So if you make a line perpendicular to this, let's say by construction, I'm trying to make this diagram, let's say, let's say I know my P, I know my A, I know my B. So let's say I know this. Now I draw a line perpendicular to this. So let's say the line goes like this. Similarly, if I draw a line perpendicular to this and it goes like this, you'll realize that they both will be basically meeting on the circle and at the center of the red circle. And therefore, you can claim that TO is the diameter. Another way of looking at it. All right. Next question that comes normally is the angle between the tangents, okay? The second type of question that comes is angle between the tangents. Let's call first of a length, length of the chord of contact. This is another type of question that comes length of the chord of contact. So let me recall the figure over here. I don't want to draw it again and again. Yes, so if I want to find out the length of the chord of contact, how will I find out? What is the length of the chord of contact? Now you can use your radius of the circle, okay? You can use the radius of the circle to find the length of the chord of contact. You can also use the length of the tangent, okay? So equation of the circle is known, points are known. So you know your center, you know your radius, you know the length of the tangent. So all these aspects can be used to get the length, okay? So let us go for the length. I would request you first of all to assume that the length of this particular tangent is let's say S. Let me not use S, let me use a K, okay? So length of the tangent is K. Radius of the circle is R. Radius of the circle is R. Tell me what is the length in terms of K and R? Tell me the length of the chord of contact in terms of K and R. I hope my question is clear to everybody. Use some ingredients of trigonometry also. Why not? You definitely know the length of the tangent, my dear Aditya, under root S1, right? So you know the circle equation, you know the point, S1 can be found out, under root S1 is the length of the tangent. So assuming that that is K, okay? So K is actually under root of S1. You all know this. If you know this, if you know the let's say R and if you know the point, you can easily get that. Yes, of course you can easily get that because you can use your Pythagoras theorem for that. In fact, Pythagoras theorem itself was used to derive this formula. Yes, anybody with the expression? If you have done anything, it is too big for you to type out, you can also say done. But if you think you can type it out, it is not that ugly an expression, then go ahead, type it out. Okay, Kinshuk has already given an answer. Very good, Kinshuk. Yes, yes, Kinshuk, I got your answer. Yes, okay, so let's discuss it. It is not a big deal, it is not a big deal. We all know from our class 10th days that these two triangles are congruent. Okay, so this angle and this angle are same. Let's say I call this angle as a theta angle. Okay, now if you look at triangle P O A, can I say tan of theta is R by K, right? Yes or no? And we also know that this is going to be half the chord because any perpendicular line from the center bisects the chord, okay. Can I say the same theta, tan of theta, can also be, sorry, can I say the same theta, that is sine of theta, can also be used to write L by 2K, yes or no? Correct? Now, if sine theta is L by 2K and tan theta is R by K, can we eliminate the theta between them and get a value of L in terms of R and K? Okay, so you'll say very simple. So how do you eliminate theta? There's so many things we can use. We can use our identities, correct? Or you can directly use the fact that if tan theta is this, then sine theta should have been R by under root of R square plus K square, correct? So now comparing these two results, L by 2K is equal to R by under root of R square plus K square. Am I right? Am I right? Any questions? So from here, I can say your L value will be 2KR by under root of R square plus K square, correct? Now again, I'm not emphasizing on you remembering this result. If you can basically find it out from the real time, that means by using the numbers and figures given to you in the question, well-input. Anyways, there's so many ways also to find it out. Is it fine? Okay, the next type of question that comes, I'm so sorry, why should you get a 2K? Okay, see here, Siddhanth, this length is completely L. So if I take this part where the perpendicular from the center is hitting this chord of contact, it will divide it into L by 2, L by 2. It bisects the chord, right? It's a perpendicular bisector to the chord. So now this is 90 degree and this is L by 2. So basically using the same angle theta, I wrote a trigonometric ratio opposite by hypotenuse is sine theta. Opposite is L by 2. So L by 2 pi K will become L by 2 pi K, so L by 2 K. Make sense, Siddhanth? Happy? Okay, cool. Next question that normally comes is, what is the angle between the two tangents? So let's take it as a third thing. So what is the angle between, in fact, I will tell you another way to look at it a little later on, angle between the pair of tangents. Between the pair of tangents. Let me recall my figure yet again because I don't want to make it again. So there you go. Yeah. So let's find out, let's find out what is the angle between these two pair of tangents? So what is this angle? Let's say I call it as five. Yes. Again, use your length of the tangent, let's say K, and use the radius of the circle, okay? To give me that angle, this is a very simple exercise. You probably have already done it. Done? Everybody's done? This is just, this can be done within a blink of an eye. So you know phi has to be two theta. You know phi has to be two theta, right? And from the given diagram, theta is nothing but tan inverse of r by K. Okay, so this is going to be your answer. But normally if let's say somebody has to write it in terms of a single trigonometric ratio, okay? Then normally we write it like this. Tan inverse of two r by K divided by one minus r by K the whole square. Now how does this formula actually come? Now all of you are actually studying inverse trigonometry. Okay, so I would like you to figure this out. How does this actually come? Okay, and of course on simplification, I can write it further more like two rK by K square minus r square. Okay, so can somebody tell me, from here to here, how does this come? Now napplers would be knowing it. Any nappler here? Yeah, Pakul is there. Raghav, are you there? Raghav is also there. Okay, from tan to theta formula, excellent and you wrote the very good, so good start. So basically let us say, let us say, let us say, let us say tan inverse, I'll just give you a proof for this. Let's say tan inverse r by K is some angle. Let me call it as alpha for the time being. Okay, now what do I need? I need two alpha, right? So what I will do, I will start with the fact that tan of alpha is r by K, okay? And then I will use my identity, double angle identity of class 11, which I had learned. That is tan to alpha is two tan alpha by one minus tan square alpha, correct? And now tan alpha is r by K, so I'll put r by K in terms, in place of tan alpha. So this is what I see, okay? This is tan of two alpha. So this is nothing but two rK by K square minus r square. This is tan of two alpha, correct? Let me just put a barricade over here. And ultimately, since I need two alpha, if you see your diagram, if you see your answer over here, it is two alpha, right? So two alpha from here can be obtained by taking tan inverse and this, there it goes. This gives you the given result. Now again, this result is not that easily obtained when you look at your inverse signometry point of view because tan inverse of tan theta necessarily doesn't mean tan theta, necessarily doesn't mean theta, okay? But here in this case, all the required conditions are met, okay? Normally when do we write this as when, when your X in this case, tan of alpha lies between minus one and one, okay? In this case, the same tan alpha role is being paid by r by K and this is definitely more than minus one because it's the ratio of two positive quantities and it has to be less than one because length of the tangent will always be more than the radius of the circle. So only when this condition is met, we can write from this step to this step, okay? And here that condition is actually met. Here this condition is actually met. Now, how does this condition come? As of now, I have not taught this to you in inverse tignometry. Wait for Thursday's class for Raja Ji Nagar and others, please wait for Friday's class for this to happen, okay? Anyways, moving on, let's take a question on this. Let's take a question on this. Tangents have I done for the contact? Okay, so I think let's start with this question. Maybe you have done this in your DPP or maybe you may not have tried it in your DPP. So still, let us do this question. The question is, tangents are drawn from one comma eight to this circle, which touches the circle. One second, Abhinav, I'm sorry, Abhinav, can I go back to the slide after we are done with this question because people have started reading and solving it? If you could just hold for a few seconds. Thank you. Just remind me to go back, okay? I may forget. Yes, so tangents are drawn from external point to the circle which does the circle at A and B. The equation of the circum-circle of triangle PAB. I want this response to come within one minute and putting the poll on, okay? Please start responding. So how many of you get up in the morning and the first thing that you do is, of course, do your morning duties, take your breakfast. How many of you actually did it? To be very frank, oh, very good, Shepish. I know, I know. That includes me also. Everybody would be cursing me. Just said I skipped the class on Sunday morning when you're supposed to sleep for long. You could have at 7.55 over, just right on time. Very good, so almost 10 of you have responded and one of the options is getting the maximum number of votes. Very good. All right, last 15 seconds, everybody should respond because this is a very easy question. I've already discussed about it. Okay, five, four, three, two, one, go. Go, go, go. Okay, so basically as we discussed, most of you, by the way, have given B as your chosen option. 69% of you have chosen B. Okay, now as I told you, the circum circle would be having this point and the center of the circle as the diametrically opposite ends. So center of this circle, we all know, the center of this circle is going to be 3,2, correct? So just you have to write the equation of a circle passing through these two points. So x minus x1, x minus x2, y minus y1, y minus y2 is equal to zero. On simplification, this will give you x2 minus 4x plus three and this will give you y2 minus 10y plus 16. If you simplify it, you get x2 plus y2 minus 4x minus 10y plus 19 equal to zero, which is definitely option number B. Easy question, not an issue. Let's move on. Okay, let's take this one. I think this also you might have done in your DPP. If not, we will try it out here. The question says, oh, I'm so sorry. One second, yes, yes, yes, Abhinav wanted me to go back. Aha, see, I told Abhinav, I may forget. Yes, Abhinav, sorry. Anything that you would like to know from me or you just wanted to copy something. Let me know once you're done. Okay, thank you, thanks a lot. Let's move on to our next question. The next question is, tangents are drawn tangents are drawn from the point on this circle to another circle. And the chord of contact touches another circle. Okay, so there are three circles involved and you can see how they are related to each other. Then you have to comment upon ABC relation among each other. Are they in AP? Are they in GP? Or are they in HP? Okay, I'm assuming ABC, they all are the radius of the corresponding circle and hence their positive quantities. So let's try to solve this. Let me give you around two to and a half minutes for this. Try it out. Meanwhile, as you are trying, I'll try to make a diagram. Okay, if you want a poll, I can run the poll. I'm relaunching it, okay? Ah, sorry, I missed out on that too. Very good. I've got two people responding so far. If you cannot see the poll, if you are unable to see the poll, you may also give me a response on the chat box. Very good. Six of you have already given me the response. Well done. I'll give one more minute early. One, one more minute, one and a half more minutes. Oh, I'm so sorry. Now C square will be the inside most. Tangents are drawn from a point on the circle with radius A to the circle of radius B. Whose chord of contact has the circle of radius C? So radius C would be the interior most. Yeah. And they all will be concentrated because they all have their centers at origin. 18 of you have responded. Dear all, please hurry up. I'll give you 20 more seconds. Okay, five, four, three, two, one. Okay, so half the class has only chosen to respond. Never mind. Out of the people who have responded, 56% say option number B. That means they're in GP. Okay, let's check whether they're in GP or not. See, again, the story starts with you drawing two tangents from a point on the circle. Now, what is that point you don't know? So let's say if you have to choose a point on that circle, how would you choose it? If I have to choose this point, let's say point B, how would I choose this point? I would normally choose this point as a parametric point. Okay, this is the beauty of parametric equations that we had learned in our very first class. So if I have to take a point on this circle, I would normally take it as A cos theta comma A sin theta. Okay, it's hassle free because you just have one variable to deal with, which is theta in this case. Now, what is this squad of contact equation? To this circle, the circle drawn in yellow. So now your X1, Y1, X1, Y1 point is your A cos theta comma A sin theta, right? Equation of the chord is given as X, X1 plus Y, Y1. In this case, it'll be B square because you're drawing it to the circle having the radius B square, not to the circle having a radius A square, right? So see the concept involved here. The point is on the blue circle, but you are drawing a chord of contact to the yellow circle. So you'll find T equal to zero with respect to the yellow circle using the point X1, Y1, which lies on the blue circle. So what I do, is it making sense? Okay, so this is going to be A cos theta, A sin theta equal to B square. Okay, so basically it's nothing but X cos theta plus Y sin theta is equal to B square by A. Am I right? Now, this fellow behaves like tangent. This fellow is a tangent to the inside circle. Isn't it? This fellow is a tangent to X square plus Y square is equal to C square, correct? That means, if I use my condition of tangency or if I use my point, a fact that the distance of the center form, this line should be the radius, anyhow I'm going to get the same result. So what do you want to use? You choose your, this thing. So I can use this fact that if this is a tangent, then let's say, had I written it as Y is equal to Mx plus C, let me write it that way. So Y is equal to, or Y to make it complicated, rather let us write this down in our general form like this and use the fact that distance of this line from the origin should be equal to C. Okay, so origin is zero zero, so it'll be modulus of minus B square by A by under root of cos square plus sin square. This distance should actually be your C. Yes or no? Okay, now this is anyways one. So you can write this as B square by A is equal to C, that means B square is equal to AC, okay? They are all positive quantities, so there's no point putting a mod next to A and next to C and all. So this is the relation which A, B, C will obey and that clearly implies they must be in GP, okay? Now I have a question for all of you. Do the same problem by using just geometry, no coordinate geometry. So try out as a homework, solve the problem, solve the problem by using just geometry, no coordinate geometry, no equation and all, just distances and all, geometry, okay? And do send it to me personally, okay? This is for your homework. Yes, same dimension, Gayatri, they all have their center at the same position, okay? And one has a radius of A, another has a radius of B, another has a radius of C, okay? And of course, all these conditions that is given in the equation, that is met by these circles. So no equations, no coordinates to be used, just in geometry, try to solve this question, okay? Anyways, so today my next concept would be equation of pair of tangents. So we have been using pair of tangents, but we didn't learn about what is the equation of pair of tangents. Now this concept to a certain level would be related to pair of lines, okay? Most of you have not done pair of lines. In fact, I have not officially done this chapter with you. That is a part of my agenda, Sunday agenda. So in Sunday agenda, I'll be covering up all the conics, including pair of lines. And I also plan to do with you a theory of equations because many of you were asking me about the Cardin's formula, the Descartes rule, all those things. So we'll take those concepts also. Yes, so what is pair of straight lines? See, pair of straight lines in a plain and simple word, if I tell you. It's nothing but if you have two lines like this, okay, let's say one line is like this, another line is like this. Now they need not be intersecting, okay? They may be parallel also, okay? They may be coincident also, okay? So let's say this line is a1x plus b1y plus c1 equal to zero, and this line is a2x plus b2y plus c2 equal to zero. When I say what is the equation of the pair of straight lines, all you need to do is just write them like this. Just multiply these two equations in this form. But only to be used when you have converted them to the general form, okay? I'm so sorry, yeah. Okay, so when you multiply it like this, this becomes the equation of the pair of these straight lines, okay? So normally when you expand such a equation, you would realize that you end up getting a second degree general equation. So you'll have an x squared term, it will have a y squared term, it will have an xy term, it will just have an x term, just have a y term, just have a constant term, okay? Now it is very important to write them in the general form before multiplying it, right? Why I will tell you the reason when I do the chapter with you. So let's say, I didn't get that point. Ginshukh, so the point either lies on line one or line two. I mean, I'm not talking about the pair of tangents as of now, I'll come to it. In the case which we are going to take up, there the point lies in the intersection point, okay? I didn't actually get what you're trying to say, but let's try to solve that particular problem. So anyways, this is something which I'm going to discuss in detail later, okay? Later I will talk about it in much more detail, don't worry. There's a full-fledged chapter on pair of straight lines for you, okay? We'll talk about it. Anyways, coming to this scenario, how do I get the pair of tangents? Again, let me recall my figure, why to draw it again. I've already drawn it, yeah. So now let us start our discussion with first of all the standard form of a circle. So let's say I have a standard form of a circle, X square plus Y square is equal to A square, is equal to R square, let us say. And I want to know what is the equation of these pair of tangents, okay? So how would I find this out? All of you please pay attention to whatever I'm going to say, okay? If you know the equation of the circle, you basically know the equation of the chord of contact, correct? Equation of the chord of contact will be X, X1 plus Y, Y1 equal to R square, yes? So this equation is known, this point is already given to you, no problem, okay? Now, listen to this, this is a very important statement. Can I say the pair of tangents equation is basically locus of such points, locus of such points H comma K which satisfies the fact that, which satisfies the fact that, let's say I call this as point M and I call this point as N, which satisfies the fact that triangle, let me name it as Q for the timing, such that triangle QAM and PAN, fun, okay? QAM and PAN, they are similar. Are you convinced with that statement? So again, I'll repeat my statement. The pair of tangent equation is locus of such points, locus of such points, such that if I draw these two kind of figures, means if I drop perpendicular from those points onto the chord of contact, I will get two such triangles which will be similar to each other, isn't it? Everybody agrees with it? Do you want me to repeat my statement once again or have you heard it properly? Okay, so if you heard it properly, let's go ahead and use exploit that situation, okay? I'll repeat again, I need to go on repeat again, okay? So can I say a pair of tangent equation is locus of all such points, Q says that if you draw two triangles like this shown to you on this figure, the two triangles will be similar to each other. That means triangle QAM will be similar to triangle PAN. Okay, this anybody can intuitively figure out. One, if you take here, will it violate the situation? Anusha, had I taken my Q here, will it violate it? No, all right, if I take here, will it violate it? If I take here, will it violate it? No, okay, anyways, let's use this scenario. Now in this scenario, I'm going to use my proportionality here. Now all of you please pay attention. Can I say QM by PN, QM by PN will be PA by QA, oh sorry, QA by PA, other way around, okay? So if they are similar, I can say QM by PN should be equal to QA by PA, basic proportionality curve, correct? Now, what is QM? QM is nothing but the perpendicular distance of H comma K from COC, correct? So QM is what, QM is the perpendicular distance, let me write it over here. QM is the perpendicular distance of H comma K from the COC. So what are the perpendicular distance of any point from a given line? Simple, let me put my X as H, so HX1, KY1 minus R square mod by under root of X1 square Y1 square. This is QM, this whole thing divided by, this whole thing divided by, let me choose a line to do that, yeah. So this whole thing divided by PN, now what is PN? PN is the perpendicular distance of P from the COC. So what are the perpendicular distance of X1, Y1 from the COC? So all you would do is mod, replace your X1 with X1 itself, so it will become X1 square, this will become Y1 square minus R square by under root of again X1 square so eventually this X1 square, Y1 square under root and this X1 square, Y1 square under root will get cancelled off, correct? Now this is equal to, this is equal to QA. Now what is the length of a tangent drawn from an external point to a circle under root S1? You all know that, correct? So now your point, external point is H comma K and this circle is already known, X square plus Y square equal to R square. So you just have to write under root S1 over here. So under root S1 will be under root of X square plus K square minus R square. Does it make sense? Does this make sense? So distance of the length of the tangent drawn from an external point on to the circle is under root of S1. Yes, no, maybe, all agree, all agree with that? Okay, and can I say the distance of the length of the tangent PA would be under root of S1 calculated with respect to P, so that will be under root of X1 square plus Y1 square minus R square, okay? So basically I have just plugged in these values on this side. Anybody has doubt related to this, please immediately stop me because I am going to simplify this expression. I'm going to simplify this expression. Anybody has any issues? Siddhant, any issues? You said yes. Yes means okay, you're agreeing with me, okay, fine. Okay, so everybody's convinced. Could you explain the PN length? See, you don't have to rely on me to find PN length. If you know a point and if you know an equation of a line, how do you find the length of a perpendicular drawn from a point to a line? The same formula. This is your line and this is your point X1, Y1. So if I ask you, give me the length of the perpendicular from that point on to this line. What will you say? You're going to replace that point in place of X and Y, isn't it? So won't it become X1 square, Y1 square minus, again, I'm writing it in the general form of a line. Modulus of that divided by square of the coefficients of X and Y under root. So square of X1 square, Y1 square under root. That is what I wrote down in the denominator here in yellow. Make sense? Okay, so see, the moment you know this, the moment you know this, you should not need my help actually. You know what to do. These ingredients are well known to you. How to find the length of the perpendicular, how to find out the length of the tangent that everything is drawn to you. So based on that, I got this result, okay? So now let us simplify it. So on simplification, I can do one thing. I can cancel off this guy with this guy. So it will lead to, it will lead to mod of, mod of, in fact, let us square also both the sides so that we get rid of mod and all, and we get rid of the square root. So this square by this square is equal to this by this. And anyways, you see this expression is very similar to this. So one of the powers will also get cancelled off. Correct? Now let us take this part to the other side. So it will be hx1, ky1 minus r square whole square is equal to s square plus k square minus r square times x1 square y1 square minus r square, okay? Now here we need to generalize, right? So this was with respect to a very specific position that we had written this. Now we need to generalize. So when we generalize, what do we do? We replace our h with an x and k with a y. So on doing that, you'll end up getting xx1, yy1 minus r square whole square. This will become x square plus y square minus r square. And this becomes x1 square y1 square minus r square. In fact, it doesn't change, okay? Now I would like everybody to have a very, very deep look at this particular thing. What have I written actually? Can I fit in my t and s and s1 somewhere? Can I fit my t expression, s expression, s1 expression somewhere? Just have a look at it, okay? If yes, can I fit my t, s, s, s1 somewhere? Please let me know where all. Let me take a snapshot of this. Ayurama. Normally when we write, we write it all crooked. Okay, so let me just take this to the side. I hope you have copied this down because then people will say, sir, can you move to the left, move to the right, move up, move down. Done, copied. Are you done? Okay, dear all, one more important thing. Even though we will complete our syllabus by November, December period, the classes will continue. Let's say if you are a G-advance aspirant, it will continue all the way till June, July, okay? Are you getting my point? If at all the G-advance exam happens so late, okay? Normally it should happen by May, but this time, I don't know, they have kept July and that is still subject to postponement. So don't be in a rush to understand things in a very hasty manner, okay? Syllabus will be covered, but we'll do a lot of problem practice. Now you are seniors, I'm doing advanced level questions with them, okay? Same thing, advanced level questions. So whatever is your backlog, our arms are completed. Don't be like, oh my God, I'm already late, I've not done my 11th, then don't start panicking. It not serve the purpose, solve the purpose, okay? So do it, I mean things are right now very hazy, but of course, do not procrastinate it. So Sarah said it will get postponed, so let me start in month of December and all, okay? So there was one panic student who called me yesterday, sir, I messed up my 11th, I didn't read anything, I don't know, basic ABC of 11th. There's no need to panic, there's nothing to panic. Dire Dire, make a plan and work towards it. Okay, if you need my help guidance, we are always available. Yes, so chalo, I think Anusha has given the answer to this. Anusha has recognized that what I have written is actually, actually let me write it again, not this guy, not this guy. Let me make it slightly bigger, this part I'm raising it. Yeah, so if you see here very, very carefully, what I have actually written, what I have actually written for you on the screen is, this expression was for T, correct? So I have written T square, I'm writing it in big and bold, okay? This expression was for S, so this is S, and this expression was for S1, so this is S1. So here is another formula that you need to update your formula list with. So this is the formula of the pair of tangents. This is a formula for the pair of tangents drawn from an external point onto a given circle. I'll box it out so that you remember it. So T, S, S1, they help us to basically remember these ugly formulas, right? So if you know what's T, if you know what's S, if you know what's S1, all you need to remember is T square is equal to S1, that's it. This entire equation will actually give you the pair of tangent equation, okay? Again, I would like to remind you T expression is a linear expression in X and Y. S is a general second-degree expression in X and Y. And S1 is a number, this is a value, a number will come out of here, five, six, hundred, minus, whatever, of course a positive number will come out because the point is exterior to the circle, okay? So some number will come out from here, okay? So when you simplify this, you will again get a second-degree general equation and that would correspond to the pair of tangents. Is it fine? Is this clear? Oh, no problem, Mishra. You can always watch the recording there. I feel so sorry to keep the class so early. Asha, one, I think Prabhasar will tell you he also wants to take some extra classes. So, so he may use your Sunday, not every time, Q Sundays. What, sir, you taking double-double classes for us? Okay, anyways, is this clear to you? Now, the benefit of this formula is you can scale it up to general form of a circle also. See, what did I derive here? I derived it for the standard form of a circle, right? So the whole thing was derived for a standard form of a circle because standard form are easy to deal with. Nanna, Munna, they are, okay, they're very easy to deal with. The general form is complicated. So I cannot do all these derivations in the general form because the figures will look ugly, right? So when we derive it, we use a standard form and then we see what is the pattern hidden in it and by that pattern, we generalize it to other form of a circle, which is the general form of a circle. So if you have been given any circle in general, don't panic, find T for it. You know how to find T. Do you know how to find T? Yes, Aditya, it'll work for every, can be used for any circle. This can be used for, used for any circle. Or you can be used for general form of a circle also. Okay? So you know your circle equation. If you know the point and somebody says, can you give me the pair of tangents? You don't have to panic. Just recall this formula. T square is equal to SS1. T, everybody knows how to find out. Replace X square with X, X1. Replace Y square with Y, Y1. Replace X with X plus X1 by two. Replace Y with Y plus Y1 by two. Replace C with C. C will not be replaced. Okay? So know your S. Yes, subcomalumen. For finding S, again, I will repeat. Make the coefficient of X square and Y square as one. It should be no other term other than one. People are still asking the same, doing the same mistake in assignments. And S1 is basically in that S expression, put the point X1, Y1. You'll get a positive number. So just fit in this formula, you are done. Now, let us try to understand the nuances of this. This looks simple, but it is not easy. What, sir? Simple and easy, yeah. Simple things may not be easy. Now, anyways, let's talk about that scenario, where if the question setter asked me the two tangents separately, what to do? Because this gives you a combined equation, isn't it? So if I am asked to write down the equation of the tangent separately, right? Italy separate, summer separate, then what will you do? Not mixing. Okay, so this is a mixed. So if you want separate, what to do? Will you sit and resolve for the linear factors on this pair of straight lines? That would be, okay, that can be done, but I'm not saying it cannot be done, but that would take a bit of your time, okay? So let us try to understand those situations through problems, okay? Let me begin with a simple problem. Let's start with a problem. Okay, we'll take this question, very simple question. This is basically a standard form, so it is the simplest of all. Find the equation of the tangents. Let's correct this question a bit, okay? Find the equation of these tangents from this point to this circle. I hope this point is outside the circle. You can actually check. S1 will be positive. Yes, it'll be one, so it'll be positive. So it's outside the circle. Everybody, please do this problem. Please do this problem and just sit down. You don't need to type your answer down because it may be a lengthy one. Yes, so every Sunday morning, 8 to 9.30, there would be a class, okay? Till we officially complete all our, you know, things to satisfaction. In case I want to do some problem solving session also, then also I can call you on a Sunday. So please Sunday, 8 o'clock, 7.30 alarm, everybody put on your phone, a recurring alarm. Okay, done, Kinshukh, done, Anusha, very good. Pradyum, done. Shruti is done, good. Aniruddh, done. Gurman, Gaurav, very good. So, chalo, we'll directly use our formula T, square is S, S1, that's it, okay? So what is T for in this case? Now remember, T is basically when you replace your X square with X, X1, Y square with Y, Y1, and of course, bring the four on the other side. This is your T, my dear. T is not an equation, T is not an equation, not an equation, I'm again telling you, T is a linear expression, okay? Zero mat likh nahi aap hai, teega. So T is a linear expression, square it. Is equal to S, S for this circle would be this. Again, this is an expression, okay? S1, S1 is when you put the points in this S formula. So it's one square plus two square minus four, okay? So basically it gives you X plus two Y minus four, the whole square is equal to X square plus Y square minus four. I think this is a one itself, no need to worry about it. And when you expand it, you end up getting X square plus four Y square plus 16 plus four X Y minus, minus 16 Y minus eight X is equal to X square plus Y square minus four, okay? Try to get rid of the terms which you can cancel, okay? And this will give you three Y square, okay? I'm just canceling, cancelling doesn't mean it's exactly getting canceling, means I have taken care of it. And you'll have four X Y, you'll have minus eight X, minus 16 Y, and the constant will be plus 20, okay? So this would be your pair of tangents, equation of pair of tangents. Is this what everybody has got? Anybody has got a different answer? Does it match? Same answer Gayatri? Okay, excellent. Now, now, let me show you another way to approach the same question, okay? Now, method number two. I'm solving the same question, but without knowing any kind of a formula for it, okay? So I know there is a circle, there is a circle whose center is at origin, okay? A radius is also known to me, radius is two, okay? So this is a circle with radius of two, okay? And I'm trying to draw a pair of tangents to this circle from an external point. And I want to know the combined equation. So let me tell you a method to get the separate equations first of all, okay? Now, separate equation can easily be obtained by using your basics. Now, one of the basic you'll say, let the tangent equation be, let the tangent equation be Y minus two is equal to MX minus one. So you know that both of these tangents are going to pass through one comma two. So what I did, I chose an arbitrary slope, let's say M whose value is not known to me. And I made a tangent and made an equation of a line from it. Now I will use the fact that this line's distance or this line is a tangent to this particular circle, okay? That would give me a quadratic in M. That will definitely give me a quadratic in M with two real distinct roots, okay? Let's check and find that out. So if you see this, it is actually Y is equal to MX plus two minus M, correct? Now, if this is a tangent, if this is a tangent to this circle, X square plus Y square is equal to four. What is the condition of tangency? Condition of tangency, if you recall, it is C square is equal to R square one plus M square, correct? So C square in this case is this guy. This is your C for you, right? Is equal to R square, R square is this guy, correct? One plus M square. If you see here, you will actually end up getting a quadratic in M. Let us try to solve for it, okay? So clearly seeing here that you get a three M square is equal to minus four M, right? So this gives you M value as a zero or M value as a minus four by three, okay? So when you know these two M values, nobody can stop you from getting your tangency equation separately, separately. So here, if you put your M as zero, let us see what happens. When you put M as zero, you get Y minus two equal to zero as the equation of your, one of the tangents, correct? And if you put your M as minus four by three, M as minus four by three, you will end up getting Y minus two, let me write it in yellow only. Y minus two is equal to minus four by three X minus one. Okay, let's try to simplify it to a certain extent. This will give you three Y minus six is equal to minus four X plus four. So this is three Y plus four X minus 10 equal to zero. So these two are the tangent equations that you have bought separately now. Are you getting my point? Okay, now I'll tell you the method how to separate the tangent equations from the pair of tangent equation you have bought. So let us now try to combine these two equations and see whether does it give me the same equation that I have bought over here. So let's try to multiply them. I hope I'm copying it correctly. Y minus two. And the other one was, yeah, three Y plus four X minus 10. Three Y, three Y, plus four X minus 10. Okay, let's try to multiply it. So let's multiply Y's first. So you get three Y square plus four X Y minus 10 Y. Then multiply this, so you get minus six Y, minus eight X plus 20. Oh, wonderful. Three Y square, four X Y minus 16 Y minus eight X plus 20. Same to same equation I have got. Same to same equation, correct? Now what is the benefit of this method? This method actually tells you the equation of the tangent separately. The word here is separately. Make sense? Okay, so many times the question setter will give you what are the equations of the tangents drawn from this external point to that given circle? So I would recommend going with this method, okay? But let us say somebody wants to split the equation of the tangent from this equation. T square is equal to X X one. Is there any way to do that? Yes, there's a way to do that. I will run you through the process. Meanwhile, if you have any questions, if you want to copy down anything, please do let me know. Yes, that is what I'm going to do now. In short, I'm going to factor out the combined equation to get the equation of the tangent separately. Separate, separate. Yes, the second method which I've discussed will work for any type of circle. General circle, standard circle. Okay, convince, okay. Chalo, we will now discuss how to split. Splitting, how do we split a general second degree equation provided it should represent a pair of straight lines? How do we split them? So what I'm going to do is I'm just going to copy this guy to the next page because this page is completely full. So how to split this in two pair of straight lines? So I want to split this as two lines. Line L one and line L two. Anyways, let's try to understand this. So the process is step number one, you identify the second degree terms in this equation. You identify these terms. So these are the second degree terms in this equation. And do a simple, you know, factorization of this. So if I factorize this, I'll get y and three y plus four x equal to zero. Am I right? Am I right? Okay. So this basically tells you that one of the factors contains a y and other factor actually contains three y plus four x. Okay. Now, there could be constants here. Let me write L and M because the other terms would not come into picture without these constants. So had there been only y and only three y plus four x, there would have been only the second degree terms. But that is not the case. In our given expression, I have a linear term in x and y and I have a constant term also. So that could only arise when there were constants present over here. Okay. Don't equate it to zero. Right? Don't equate it to zero. Okay. So normally factorize it like this. Happy? Aditya? Okay. The next thing is I need to know my L and M because without that, my problem will not be solved. Right? So what do I do? I use the coefficient of x and y to get the value of L and M. Okay. Let me show you. So find out what is the coefficient of x from this given equation here. Oh, sorry. From this given expression over here. So you'll see x will come when L multiplies with four x. So you get four L, correct? That's the only way you can end up getting x. Okay. And what is the coefficient of x in this expression? That is minus eight, compare it. So your L will become minus two. Okay. In a similar way, compare the coefficient of y between these two expressions. So what are the coefficient of y in the second expression which I've written here? So y into M will give me a y, so I'll get an M. And three y into L will give me a y, so I'll get a three L. And compare it with the coefficient of y over here which is minus 16. I hope this is making sense. So what am I doing? I'm doing nothing great. I'm not doing any rocket science concept. I'm just picking out what is the coefficient of y had I multiplied this, okay? And don't actually multiply it. You can hand pick it just like the way I did it. And compare with the coefficient of y of this equation. Okay. And solve for these two. So here you have already solved for L. So let's solve for M. Oh, I'm so sorry. So M becomes minus 10, correct? So L is known, M is known. And that's all we wanted to know because if I know that I can replace this guy with a minus two. I can replace this with a minus 10. So these are the two equations. So basically you put them individually to zero. That would be the two equations of those two tangent separately. So y minus two equal to zero and three y plus four x minus 10 equal to zero. These are your tangent equations. Is this fine everybody? Is are you all happy? Gaurav, Gayatri, Gurman, Kiran, Akansha, clear as of which, Rashmika, very good. Sir, by factorizing the second degree terms, how will we show that they will be in L1 and L2? See, normally I'll tell you something very interesting. If you see your second degree terms in the equation of a line, let me write it as two h x y plus b y square plus two g x plus two f y plus c equal to zero. Now, when these lines are actually made up of two lines, let's say I call one of the lines as a one x plus b one y plus c one and a two x plus b two y plus c two equal to zero. If you know that these are made up of two lines, then these second degree terms would be made up of interaction of these two lines from each other. For example, this x square term would be obtained only when these two guys interact. This y square term will be obtained only when these two guys interact. And this two x y term will be obtained when this interacts with the other one and this interacts with this one. There's a cross-linkage interaction, right? The role of c one and c two is only to produce these last three terms, right? So let us say, had I dropped these terms, then these c one and c two would not have come into picture. Right? Aditya? So what I did was, I did the opposite way round. I took this term and factorized it like this. I would only get these two expressions. Yes or no? So the only two linear factors would be multiplying to give me these second degree terms only. That is what precisely I used. Normally I do this exercise in the pair of state lines chapter. So when we do that chapter, we are going to revisit this concept yet again. All right. So meanwhile, we'll reinforce our concepts with more questions. So let's take few more questions. So I hope the splitting of the second degree into linear factors is clear to everybody. Okay, now many people ask me this question. Sir, what if this was not factorizable? Then use your Shridharacharya formula. What if I don't get real roots? If you don't get real roots, that means there is no real, no tangents existing because the slopes will be non real. Okay. So all those aspects I'm going to cover under pair of state lines, not to worry right now. Right now we'll worry about solving few more questions. No, no, no, no, no, this one, this one, yeah. Okay. Now we will do this question in two ways. One is by of course our regular formula because right now, as you can see in your options, they have combined equation of pair of tangents, but we'll also try to get the same combined equation by individually finding their tangents and multiplying it. I think we have a good amount of time to complete this exercise. So we'll do that. Let me put the poll also. If you're done, you can respond on the poll. So this Thursday, your schools are going to reopen, right? Most of your schools are going to reopen Thursday. June 2nd, Gauravka June 2nd, okay. June 2nd is what? 21st is Monday, one's Wednesday, okay. Normally NPS opens on a guru war. Guru means master, I mean the teacher. So that's why they only reopen on Thursdays. Okay, NPS, Yashwantpur has already opened. Okay. So Pushplata ma'am is there, taking maths in one section. So who's taking for your classes? Okay, Sabitha ma'am. Seven of you have already responded. Dear all, let's finish it off in next two minutes. So next class we'll be starting with chord bisected at a point and we have to also do director circle and family of, sorry, common tangents, intersection of circles, condition of orthogonality, family of circles, equation of the diameter, concept of radical axis, common chord, pole and polar. So I think three more Sundays we'll go. See that's the reason why I did, I could not complete this topic in 11 because I don't want to do a half-hearted effort, okay. It's very easy for any teacher to just skip a lot of topics and declare that this chapter is over. No, I don't do that. I tell you more than what is required actually and that really helps. Okay, the rank one who came this year in JMA, he said that those extra concepts really shortens his, time of solving questions. So the similar type of concepts would be seen in other conics. So if I do circle in detail long, where I will hyperbola ellipse, I can jump few concepts because the same equation T equal to zero, T square is equal to SS1, et cetera, they are there for other conics also. So I can save time there, okay. So in one topic, I have to do it in detail, absolutely. Okay, should we discuss it now? People who want to vote, please go for it. At least this problem you would have solved, you probably would be trying to solve the same thing by separately finding out the tangents. So if you have solved this question, go ahead and give your response so that we can close the poll and start discussing it. Guys, this font, this font, just 18 of you, what is this? There are 62 of you. Okay, five, four, three, two, one, go. I'll also take your attendance meanwhile. All right, so let me stop the poll now. So most of you have given a mixed response between B and C. I don't know why it was a straightforward question. Okay, so most of you have said B by the way, but people have been wrong as well. I mean, I'm not sure whether B is right or not. Let's look into this. So this is a point from where you're drawing a pair of tangents, right? So basically your S, see, the best thing is write your S, S1, T, simple as that. S, it's very obvious, S will be your X square plus Y square minus two X plus four Y plus three. S1 will be when you're putting the point in place of X and Y. So this will be six square. Let me not write minus five square. And this will be minus two X minus two into six plus four into minus five plus three. Okay, let's calculate this number. So 36 plus 25 minus 12 minus 20 plus three. So this is going to give you 61, 61 minus 32 plus three. 64 minus 32. 32? Are you getting 32 for this? Okay. T1, sorry, T. There's nothing called T1, T. T is obtained by putting this, you know, using X, X1, X, X1, okay? Y, Y1, okay? And replace your, replace your X with X plus X1 by two. Replace your Y with Y plus Y1 by two. Let the constant be as it is. Okay? So let me simplify this. So this will give us, this will give us six X minus five Y minus X minus six plus two Y minus 10 plus three, okay? On simplification further, this should give me a five X minus three Y and this will give me minus 13, okay? So this is your T. Now, coming to the pair of tangent equation, we know it's T square is equal to SS1, okay? So T square is, let me write it in different color so that, yeah, gray color. So T square is equal to SS is your given expression, okay? S1 is 32, S1 is 32 into 32, okay? Let us try to simplify things as we solve it. So 25 X square will be generated and 32 X square will be generated over here. So can I say I will end up getting a seven X square, okay? Nine Y square will be generated and this is 32 Y square. So can I say 23 Y square will be generated? So, oh, all my options have this. So I cannot rule out my options. Now, let's talk about terms containing X, Y. Term containing X, Y will only come from this position. So it will come with minus 30 X, Y. So bring it to the other side and it will become plus 30 X, Y. So now I can rule out at least this guy. Town, town, okay? B is gone. A, B. One second, why did people choose B? Oh my God, most of you went wrong there, there itself. Are you a drama? Okay, anyways, now let's focus on X term, X term. Maybe you can choose, yes. Let's focus on X term. So the X term that will come from here is minus 130. If I'm not mistaken, minus 130. And the X term here will be minus 64. So minus 64 plus 130. How much does it give you? 66, correct? So you'll end up getting 66 X, okay? So is there anything that I can rule out from here itself? Plus 66 is only in option A. So yeah, yeah. Okay, so option A may be correct because none of these is also sitting here. None of these, yeah, this makes our life more miserable. Okay, now let's check Y. So Y from here will be, here will be 128. And from here I will get Y as minus, plus 78. So minus 78 will be, right? Sorry, 78, yeah, 78. Okay, so minus 78 will give you a 50. Is there any 50 sitting? Yes, so there is a 50 sitting over here, okay? And of course the constant terms will be 96 coming from here and minus 169 coming from here. So this is going to give you a 70, 73. 73, yeah, it's there. So this is going to be your equation of the pair of tangents. Now this is clearly option number A. Okay, option number A is correct. Now, just to clarify the doubt, can I do the same question by finding out the tangent separately and multiplying it? In this case, I would not suggest you that method because here they have given you an option in terms of combined. But let's say had the question being, what are the two tangents that you can draw, then you can use the second method. So what I'm going to do, I'm going to do a similar problem quickly by use of second method. I'll not completely do it, clearly I'll just initiate the question. I'll just, let me just put the question once again. So what I'm going to do is, first I'm going to figure out what is the center and radius of the circle? So center of the circle is at one comma minus two. What is the radius of the circle? Radius of the circle will be under root one square minus two square minus three. What is it? That comes out to be a root of two. Okay. Now, I will assume that, let the tangent equation, let the tangent equation be y minus of minus five is equal to mx minus six. In other words, I'm assuming the tangent equation to be y is equal to mx minus six m minus five. Okay, better to write it on one side like this. Okay, let this be the tangent. This is the tangent. That means the distance of the center from this should match with the radius, correct? So what are the distance of the center? It will be mod of minus two minus m plus six m plus five by under root of one plus m square. This should be equal to the radius of the circle. Okay. Now, if you look at this term, this is actually five m plus three mod is equal to under root of two times one plus m square. Correct? Square both the sides. Square both the sides. When you square both the sides, you'll end up getting five m plus three the whole square is equal to two times one plus m square. Okay. This simplified 25 m square plus 30 m plus nine is equal to two plus two m square. That will give you 23 m square, 23 m square, 23 m square plus 30 m plus seven. Okay. Now, as you can see, this is very much similar to the expression that we had in our equation of the circle. Okay. Check the equation of the circle. Basically, if you divide your second degree terms, sorry, if you divide your second degree terms over here, equation of the pair of tangents, if you divide this by x square, you will end up getting something like this. Seven plus 23 y by x the whole square and 30 y by x replace your y by x with m. Okay. So this will give you 23 m square plus 30 m. This is exactly the same thing that you have seen over here. Okay. So again, go for the simplification factorize it. This is very easily factorizable. Okay. This can be broken as 23 m plus seven m take 23 common. 23 m common. Okay. So 23 m plus seven and m plus one equal to zero. So this results into two values. M is minus 20, minus seven by 23 and m is equal to minus one. So use this, use this over here in this equation or this equation which you have initially formulated. So the two tangents that you should be getting would be of this nature. Y plus five is equal to minus seven by 23 x minus six. This is one of the answers. And Y plus five is equal to minus one x minus six. This is another answer. So just multiply it, you'll get the same equation back. Okay. So I think we have covered up this concept pair of tangent equations to our satisfaction. So next class when we meet, we are going to take up next Sunday, which when we are going to meet again for circles, I'm going to take up other concepts. So I think we are going by two to three concepts per class. Okay. All right. Thank you. Thanks a lot. Take care and stay safe. Bye-bye.