 So I'll start my today's session with the last question that we did in the previous circle session because there was a confusion with respect to the transverse common tangents. Okay, and I'll take that question once again for you, just for the transverse common tangents and post that I will talk about angle of intersection of two circles and I will also touch upon the concept of pole and polar today because in some of the DPPs, the concept of pole and polars are there. However, pole and polar is not a very, very important concept. I mean, in case J advance decides to ask it, it will ask it as a comprehension type question. Okay, so we'll be prepared with everything. Don't worry. So let's take the last question of previous session. The question was find all common tangents, but we'll find only the transverse common tangents. Find transverse common, common tangents for the equation x square plus y square plus 6x minus 2y plus 1 equal to 0 and for the equation x square plus y square minus 2x minus 6y plus 9 equal to 0. Okay, we had figured out the center of this circle. The center of this circle was at minus 3 comma 1 and this was the center. Let me call it as C1 and the radius of this circle, the radius of this circle was 3 units. Okay, so I'll just quickly write down that information because I don't want to spend time there and the radius of this guy was 1 unit. Okay, and we were looking for the transverse common tangents for these two circles. So let me make the two circles first of all. Okay, let me make them spaced apart again. So let's say this is your C1 minus 3 comma 1. This is your C2 1 comma 3. The radius of this was 3 units. The radius of this was 1 unit. So you can see the larger one has a 3 unit. Now I want to know the equation of the transverse common tangent. Now we all know that the transverse common tangents and the line connecting the centers of the circle meet at ICS. What is ICS? Internal Center of Simulitude. So if you connect this, it will meet at this point T and basically T divides the joint of C1, C2 internally in the ratio of 3 is to 1. Okay, so first let us figure out what is the coordinate of T. So for coordinates of T, you can use your section formula. Internal division section formula. So 3 into 1, 3 into 1, okay. 1 into minus 3, 1 into minus 3 by 3 plus 1 comma 3 into 3 plus 1 into 1 by 3 plus 1. So this clearly gives you 0 comma. This will be 9 plus 1, 10 divided by 4. 10 divided by 4 is 5 by 2. Correct? No, I'm not talking about the last class one. Last class was DCT. I'm talking about DCT today. Transverse common tangent. There are two groups of tangents, Shraddha. One is direct common tangents. Direct common tangents meet at external center of simulitude. That is the point where the joint of C1, C2 is externally divided in the ratio of R1 and R2. This is transverse common tangent. Okay. So see the question. The question is already saying, find the transverse common tangents. T, C, T. Okay. Alright, so moving on with this. So having got this point. Having got this point. Now we are going to assume that there is a line with a slope of M passing through T. Okay. So let the equation of one of the TCTs be y minus 5 by 2 is Mx minus 0. Okay. So what I'm assuming, I'm assuming that the equation of the tangent, or equation of a transverse common tangent is given by this. Okay. So let's simplify this. It gives you 2 Mx plus 5. Okay. Now let me write it in the general form. Now this line, this line is a tangent to any one of the two circles. You can take any one of the two circles. Let me take the smaller one, C2. It's a tangent to C2. If this line is a tangent to C2, that means the distance, oh my bad. Yeah. That means the distance of this line from the center of C2 should be equal to 1. That is the radius. So this distance, this distance should be equal to 1. Okay. So we'll use that. So we'll put 1 comma 3 in this point and find out the distance. So it will be 2 into 3 minus 2 M into 1 minus 5 mod by under root of 2 square plus 2 M square. And this should be equal to 1. This is fine. Any questions? Any concerns here? Okay. So 1 and 3, I put it over here, 1 and 3. And then I found out that now this gives you mod of 2 M minus 1. So this is mod of 2 M minus 1. And this is equal to under root of under root of 4 plus 4 M square. Let's square both the sides. If you square both the sides, you'll end up getting 2 M minus 1. The whole square is equal to 4 M square plus 4. On simplification, if you realize, one very interesting thing is happening over here. The interesting thing is the M square term goes off. So we were supposed to get a quadratic out of it because we should be having 2 M values for the two slopes of the TCTs. But here you realize that your M square term is getting cancelled. In other words, your M is just coming out to be minus 3 by 4. Okay. So what does it actually mean? If you are getting only one M value from that particular equation, or for that matter, the quadratic equation has the coefficient of M square as 0. So if you properly write it, if you properly write it, you are writing something like the 0 M square minus 4 M minus 3 equal to 0, right? So in a quadratic, when the coefficient of M square becomes or the coefficient of X square talking about, let's say, a normal quadratic equation becomes 0, what does it indicate? It indicates that the quadratic has actually become a linear. When quadratic becomes linear means one of the roots has actually gone to plus or minus infinity. See, this is what happens. Let me just show you on the Geogebra. So I'm just going to write a simple quadratic. Yeah, one second. So first of all, I'm just going to, you know, take a value. Yeah, take a value of A and keep it to, keep it to, let's say, minus 5 to, let's say, now I'll write a quadratic using this. So Y is equal to AX square plus, let's say, and a minus, let's say 3X plus 2. Okay. Or let's say minus 10, minus, not 10, minus 6. Okay. Now if you see, this is the quadratic. Now I'm going to play with my A value. I'm going to drag this towards 0. See what will happen. When you are going towards 0, the moment you make a 0, you realize that this graph will convert to a linear. If you see this graph, this has become a linear graph of minus 3X minus 1 equal to 0. What does it indicate? One of the roots is here. Other root has gone to infinity or minus infinity. You might, you know, term it as anything. Okay. Doesn't make a difference. But what it indicates that is one of the roots is minus 3 by 4 and the other root, let's say I call it as M1, then the other root will actually be tending to infinity. Now this is a valid answer for us because in lines, we can have slopes tending to infinity. Right. So how do I write the equation here? Now see the equation. This is how we write it. So first of all, y minus 5 by 2 is equal to M. This M is minus 3 by 4. So write it down. Minus 3 by 4X. So simplify this, multiply throughout with 4. Probably this will give you something like this. Or 3X plus 4Y minus 10 equal to 0. So this is one of the answers. Okay. And for the other answer, if your M tends to infinity, it is as good as saying X minus 0 is 0. That means X equal to 0. That means your Y axis will become another answer to this question. Okay. So don't be surprised that, oh, I'm getting one of my answers as infinity and there's supposed to be two answers. It's fine. Infinity is one of the, you can say, accepted roots of this equation, especially when you're talking about the slope. Okay. Slope can tend to infinity. Okay. And now some questions, sir. Those circles have one point of intersection. Those circles have one point of intersection. No, they're not intersecting at all. Sir, can you show the diagram again? Yeah. See the diagram. So the circles are so oriented that one of the transverse common tangents is vertical. Now, this is something which is very subjective. We cannot comment on that, but especially when you're dealing with, this is something I told in the straight line chapter also. If you remember, in straight line chapter, the slope can be infinity. So if infinity comes out as an answer from solving a equation which is dealing with the slope, then don't ignore it. That could be one of the answers. Could you explain the M part again? Did you join late by any chance? Which part, this part or the solution part can joke? Which part are you talking about? Infinity thing. See, see what happens normally in our normal circumstances. This should give me a quadratic in M. But here what you're seeing is that there is no quadratic getting formed in M because the highest degree terms of M are getting cancelled. Now, it doesn't mean that you will not have two values. You still have two values technically because there has to be two transverse common tangents between them. So how does this equation actually support that cause? So the absence of one of the root basically indicates that the other root is infinity because the coefficient of M squared actually becomes a zero. So see, when the coefficient of X squared becomes zero, what can you comment about the sum of the roots or product of the roots? That will also tend to infinity. It is because one of the quantity is finite and the other quantity is infinite. So the product becomes infinite on and the sum also becomes infinite. So this is how this particular fact justifies that there could be infinity as one of the or minus infinity as one of the slopes coming out from this equation. Is this fine? I think Anusha was the one who asked this question to me as a doubt. She was trying to solve it and she realized that one of her roots was not coming up. So in such cases, please see that is why I always say drawing a diagram accurately always helps. So you will when you draw this diagram accurately, you would realize that there was actually an infinite tangent. Let me show you how. So let me just go to Jujube again. Let me make both the circles for you. So that will make more sense to you. So let me type down the equation. So first equation was X squared plus Y squared minus 2X minus 6Y minus 6Y plus 9 equal to 0 plus 9 equal to 0. And the other equation was X squared plus Y squared plus 6X plus 6X minus 2Y minus 2Y plus 1 equal to 0. Okay. You can clearly see that they have a direct, sorry, they have a transverse common tangent, which is the Y axis itself. So X equal to 0 is clearly one of the transverse common tangents. And the other one is what we figured out. What was the answer that we got? 3X, let me just drag it here. 3X plus 4Y minus 10 equal to 0. 3X plus 4Y minus 10 equal to 0. There you go. This is the other transverse common tangent. Let me just zoom in a bit. Is it visible to everybody? Okay. So with this, we conclude the direct and transverse common tangents concept. And I'm going to now talk about a concept linked to intersection of two circles. And that is the angle of intersection of two circles. So let me talk about that as well because this is one of the important concepts. Angle of intersection of two circles. Of two circles. So obviously the two circles given to you in this question must be intersecting. Okay. So let us say I take the general form. I take the general form of the two circles like this. So let's say this is your circle C1. This is your circle C2. Okay. By the way, I should refrain from using this. I should use your s and s dash as the name of the circles. So this is your another circle. So basically I'm trying to say s equal to zero and s dash equal to zero. These are the two circles given to you. Okay. And they intersect and they intersect. Let me make the two circles. So one is this and other is this. Why are you moving? Yeah. So how do you define the angle of intersection? First of all, between two circles. So angle of intersection is defined as the angle between the tangents. So let me first make tangents at the point of intersection. So I made one tangent like this. Okay. So this is a tangent to the yellow circle at this point. Let me call this point as point A. And this is a tangent. Let me make it in different color, maybe in red. This is a tangent made to the blue circle at the very same point. There you go. Okay. So how is the angle of intersection between two circles defined? Angle of intersection is defined. Let me write it down. Angle of intersection definition. This is going to be useful in your application of derivatives concept also when we are doing tangents and normals. Angles of angle of intersection between two curves is defined as the acute angle made between the tangents to the two curves. In this case, the circles at their point of intersection point of intersection. See, there are two points of intersection here. One is point A and other is point B. But if you can see from the symmetry of the figure, this angle between the two tangents or the acute angle between the two tangents. Let me call this as the acute angle. That is not going to change. So whether you make this analysis at A or make this analysis at B, theta is not going to change because of the symmetry of the two curves. Okay. So please note this definition down. It is the acute angle made between the tangents to the two curves at their point of intersection. So it is the acute angle between the tangents made to the two curves at their point of intersection. Now, how do I find this angle theta? Let's say I know everything about this circle. I know it's center. Okay. So let's say center C1 and center C2. I already know. So basically I know the distance between the centers. Tuck, tuck, tuck, tuck, tuck, tuck. And I know the respective radii also. So let's say this is your R1. This is your R2. Okay. First, let us try to find out in terms of, let's say C1, C2 is your distance B. Let's say in terms of D, R1 and R2, let us try to figure out what is this angle theta. Let us use our triangle, you know, the properties of triangle which we had studied last year under trigonometry to figure out this angle. Now, any suggestion coming from any one of you? Cosine law. Okay. How will you use cosine law? How will you use cosine law? Drop a perpendicular from where? When you say drop a perpendicular, you should mention from where and to where? From A to B. Okay. If I drop a perpendicular from A to B, okay, then how do you find it out? Sir, we know theta is angle C1, C2, A. C1, C2, C1, C2, A or C1, A, C2. C1, A, C2. C1, A, C2. Now, okay. Now, this is something very interesting. Everybody, please pay attention. Aditya is claiming that theta is this angle. Is he right or wrong? Chitraji is saying absolutely wrong. Aditya, that angle is not theta. It gives you a feeling that it is, it is not theta. Now, everybody pay attention, okay. Everybody pay, please pay attention. Now, let us say the angle which he has called as theta, I will not call it as theta because it's not theta. Let me call it as an alpha. Okay. You can call it alpha. Okay. Now, let me call this angle as 90 degree. What call, sir? It is 90 degree. Let me call this angle as 90 degree. Sorry, it is 90 degree. Okay. Now, everybody please pay attention. Right in front of, right in front of is absolutely correct. In front of is on fire. Yes. So, all of you please pay attention. If you see theta plus 90 plus alpha plus 90, right? We all know it's going to make up a 360 degree because full circle is going to get formed. Okay. So, from here can I say, correct me if I'm wrong, theta is going to be 180 minus alpha or for that matter, alpha is going to be 180 degree minus theta. Is this relation clear to everybody? Right? Now, using alpha, I can use my cosine law. That cosine law part everybody got it right. Okay. But when you're finding cosine law, you're finding it on alpha. Okay. So, cosine law says cosine of alpha is d square minus r1 square minus r2 square by, oh, sorry, one second. Other way around, I actually took a negative of it. Yeah. So, this is what the cosine law says. So, please note cosine law for anybody who has forgotten what is the cosine law. The sine law says, let's say I call this as an angle A. All this B. Cosine of angle A is nothing but B square plus C square minus A square by 2 B C. Okay. So, please recall this. So, here I can say, cos of alpha is this. Now, cos alpha is cos 180 degree minus theta. Right? And actually, this is what was going on in my mind. So, this is nothing but negative cos theta. Okay. And just send the negative sign to the other side. If you send the negative sign to the other side, you end up getting the result as d square minus r1 square minus r2 square. By 2 r1 r2. So, please, please, please make a note of this. This is a very important formula that is going to be useful. This is the angle between. This is the angle between. Okay. One important thing is, please note that you have to mention the acute angle. So, you have to take a modulus of this. Okay. So, please note. This is the angle between the two circles. And we always state the acute angle. That is why we are taking modulus of it. So, angle between two curves is always defined as the acute angle between them. Fine. Now, a couple of things that I would like to... Yes. Sure. Sure. Sure. A couple of things I would like to analyze from that formula. So, I'm waiting for Shritesh to copy anything if you want. Copy anything. Copy anything. Copy anything. Copy anything. Copy anything. Copy anything. Copy anything if you want. Shritesh, done with it? Let me know. Done. Okay. Okay. So, now a couple of things that we can figure out from this formula. Number one, if theta happens to be 90 degree, you would realize that, you would realize that cos of theta would actually become a zero. And that means d square minus r1 square minus r2 square will actually become a zero. That means r1 square plus r2 square will actually become a d square. What does this mean? It means r1, r2 and d will form a right angle triangle, thereby following a Pythagorean triplet over there. So, see what is going to happen exactly in the figure. Let me show you. So, let's say this was one of the circles and this was another circle. When this angle is 90 degree, you would realize that, please pay attention, you would realize that this tangent, the tangent drawn to the white circle and the tangent drawn to the red circle, they will eventually pass through each other's center. So, this will pass through the center of the red one and this will pass through the center of the white one. So, when 90 degree is getting formed, you realize that r1, r2 and d will form a Pythagorean triplet. That means a right angle triangle will get formed. Otherwise, it will not. Only when under a special situation, when the circles are making or when the theta angle is 90 degree, then r1, r2 and d will form a right angle triangle, suggesting the fact that this white tangent will pass through C1 and this yellow tangent will pass through C2. Why can't r1 and r2 be the tangents of the opposite circle? This is a special condition under which it will happen. Not the other way around. So, if you see this case, if you see this case, then this grey line did not pass through C1 or should not pass through C1 and this red tangent should not pass through C2. Only when the angle of intersection becomes such that theta is 90 degree, then this grey line will start passing through C1 and this red line will start passing through C2. Please make a note of this. Why? Because many people always when they are drawing the intersection, they will make the tangent pass through the center of the other. No. That is a very special occasion when it happens. When the sun and the moon and the stars are at a certain position, this will happen. Okay. So, in this case, A will be equal to theta, right? In this case, alpha will be equal to theta, right? Because as you can see in this relation, alpha is 180 degree minus theta. So, theta, if you put 90, alpha will also become 90. So, that is the situation when what Aditya said would have made sense. Then alpha and theta would have been the same, right? So, done. Good. I like the way you people are all analyzing things. That's very important, especially. Okay. Now, this is the condition where we say these circles are orthogonal to each other. Right? So, this is the condition when theta is 90. We say the circles cut each other. Orthogonally. Orthogonally. Okay. And this condition that you see over here, that is called the condition of orthogonality. Okay. But I will write this in a more robust manner in terms of G1, G2, C1 and C2 because you will say, sir has given this equation, sir is not using only. I will use it. Now, the time has come that I will use this. So, I will figure out what is the relationship between G1 and C2 because you will say, sir has given this equation, sir is not using only. So, the relationship between G1, G2, F1, F2, C1, C2 under this particular scenario which is the condition of orthogonality. Okay. So, give me one more minute. I will just derive the condition and that condition also has to be kept in mind. Very, very useful condition. Use several times in problem solving. So, what is this condition? What is this condition? Okay. Let me write it on top. Okay. Now, let me just talk about condition of orthogonality here. Condition of orthogonality. So, condition of orthogonality was R1 square plus R2 square is equal to D square. That means R1, R2 and the distance between the center must form Pythagorean interprets. Now, what is R1 square as per the equation? I don't have to even look at it. I can say directly it is G1 square plus F1 square minus C1 under root square. So, nothing will be written. And R2 square was what? G2 square plus F2 square minus C2. I hope everybody is fine with this. And what is the distance between them? Let us try to write that down. Distance between them is the distance between minus G1 minus F1 and minus G2 minus F2. So, can I say the distance between them would be given by this? Please correct me if I am wrong. Convinced? Everybody? Happy? Okay. So, let me expand the right side of it. G1 square plus G2 square minus 2 G1 G2 and this is F1 square plus F2 square minus 2 F1 F2. And on this side, we have G1 square G2 square F1 square F2 square minus C1 minus C2. Okay. So, that sounds should come of cancellation. Okay. Now, if you write it in a proper way, you will get 2 G1 G2 plus 2 F1 F2 is equal to C1 plus C2. This is something which you should note down. Put a star around it. I will show you why I am crossing it. One second. Yeah. Box. Yes. This is a very, very important condition which is called condition of orthogonality. This is called condition of orthogonality. 2 G1 G2 plus 2 F1 F2 is equal to C1 plus C2. This is the condition to be followed when two circles are intersecting in such a way that the tangents drawn to those respective circles at the point of intersection are at right angles to each other. Any questions here? Any questions here? Now, this is the one which is important. The next one is if your theta is zero, that just means the circles are touching each other. If your theta is zero, it means the circles are touching each other. The circles are touching each other. Okay. That means the tangents that you have drawn, they will become coincident. Okay. They will become coincident. That means the circles are touching each other. Can it be expressed in terms of S1 you are trying to say? I don't know. I have never tried that out but I never found any such expression also. If at all Kinshukya are able to get to that, but the question to be asked is, the moot question is, will it be helpful in some cases? Why do we normally fall on S1 t etc. Because they make our expression quite easy for us to remember. So, this expression is already easy. There is nothing like difficult about it. Both the expressions. In fact, this expression, this is the core principle and this is the outcome of that core principle. Both the expressions are quite easy. I have not tried to figure out how they would be related in terms of their respective power of points. If at all you are able to figure something out, do this close. A lot of discoveries should be kept on. You should be doing those discoveries. So, this is something which is important and with respect to this, we will take few questions now. So, just quick recap. Whatever we have done in this particular page, we basically talked about angle of intersection of two circles. Angle of intersection of two circles is defined as the acute angle between the tangents to the two circles at their point of intersection. And for that, we use cosine law. From cosine law, we came to this expression which on simplification gave you this. To make it acute, I took a theta just to be on a safer side. When theta is 90 degrees, the two circles are said to be orthogonal to each other. And the condition for orthogonality is r1 square plus r2 square is equal to d square. Or in terms of the expression of the equation, you can say 2g1g2 plus 2f1f2 is equal to c1 plus c2. Let's take questions on this. So, let's say these are the two circles. Find the angle between the two circles. See, what happens when the circles are not intersecting? If the circles are not intersecting, you will always realize that. There is no point of intersection. So, the concept of angle of intersection will not hold true. But still, if you are finding the angle using that, that means you are basically trying to find out, see, what does that angle actually help you? The angle is basically applied, let's say the circle where like this. This is your R1, this is your R2. And this is your d. In that case, I have a feeling that, I mean, I have not tried it out, but it may lead to invalid expression for cos theta. Maybe more than one, as an answer may come. I have a feeling like that. It should be actually, because if you take d to be very, very large as compared to R1 and R2. So, you can see that the effect of the R1 square and R2 square will be very less in compare. So, Pranav has confirmed my answer, he is saying that he tried it and he realized that cos theta is coming out to be more than one. Okay, thank you Pranav. He has simple question, everybody should do it quickly, quickly, quickly. You already know the formula, you just have to figure out R1, R2, d and put it in the formula. Please give me a response on the chat box. Very good, that's correct. Kinshukh, that's correct answer. See, it doesn't matter whether you're using this formula or this formula. It doesn't make a difference, because you are keeping everything under the mod situation. Ultimately, the mod is going to give you the same thing. I by 4 is absolutely correct, guys, let us discuss it. That is not a big deal. So, first of all, for the S circle, for S equal to 0 circle, the center of the circle was at 2 comma minus 3. The radius of the circle is under root of 2 square plus 9 minus 11. How much I'll give you? This gives you root 2. For S dash equal to 0, the center is going to be, the center of the circle is going to be 1 comma minus 4. So R2 is going to be under root of 1 plus 16 minus 13. So this is going to be 2, correct? So now, the distance between this, the centers is given by under root of 2 minus 1 square, which is 1. And again, this, which is 1, which is root 2. So definitely, the sum of the radii exceed the distance between the centers. So there has to be an intersection happening. And if the intersection is happening, definitely we can find out the angle of intersection by using any one of the two formula. So let us use the second one, R1 square plus R2 square minus D square by 2 root 2. So this will give us cos theta value as 4 divided by 4 root 2, which is 1 by root 2. So let's take out of the modular symbol. So theta is going to be pi by 4 or 45 degrees, whatever you want to state. So that is absolutely correct. Easy question. Arjun, how are you getting 90 dL? Check money. You got your mistake, Arjun? Okay, let's move on to the next one. This was just a warm up question for us. Next is show that the two circles, this and this cut orthogonally, show that these two circles cut orthogonally. That also is easy. You just have to verify the formula which I discussed a little while ago with you. Done. Just sit down if you're done. Okay, very simple. So what is G1 here? G1 is minus 3. What is F1 here? F1 is 2. What is C1 here? C1 is 4. What is G2 in the second equation? G2 in the second equation is half. F2 in the second equation is 2. And C2 in the second equation is 1. Now just figure it out what is 2 G1 G2 plus 2 F1 F2. So 2 G1 G2 is going to be minus 3. 2 F1 F2 is going to be 8. So the sum is coming out to be 5. And what is C1 plus C2? C1 C2 is clearly 4 plus 1 which is 5. What does this imply? This implies 2 G1 G2 plus 2 F1 F2 is equal to C1 plus C2. Implying that the circles are orthogonal to each other. Is this fine? Easy question. Not a problem. So while solving this question should be stepwise. You mean to say do we have to derive the result as, this is what you mean? Stepwise means what? Stepwise means I didn't get you. You just have to find 2 G1 G2 plus 2 F1 F2 and try to show whether it's coming out to be equal to C1 plus C2. Derivation need not be done because this is anyways, see now you're passed out 11th. This concept is anyways not going to come in any form in your board exam. If you're talking about JEE or any competitive exams, by the way, very glad to share the VIT triple results came yesterday and NPS Raja Ji Nagar student Trippan has scored all in your rank 266. Okay, it's a good rank actually because VI triple is considered to be an exam with no negative marks. So many people score good there. I mean, even people who are not that worthy, they come up in the ranking because they do some kind of a guesswork and all those things. So 266 rank, Trippan has got which is the highest that we have got. So I think Pratham got 900 something, 906 I believe, 906 rank. So Raja Ji Nagar is always as usual rocking in terms of rankings. Sir, can we also do this by finding the theta? You find R1, you find R2, you find D square and you show that R1 square plus R2 square is equal to D square. That means they form a Pythagorean triplet. From there only this condition has come to origin. This condition has not come from the sky. Okay, let's take another one. Find the equation of a circle which cuts this circle orthogonally, has its center on the line x equal to 2 and passes through 4 comma minus 1. Read the question carefully. This coming year, most of you will be registering yourself for different exams. I would say my suggestion would be register yourself at least for 7 exams, if not less. Of course, J main you will be definitely doing, CET you will definitely doing. Or CET slash Comet K depending upon your domicile criteria. Bitsat, Manipal, VIT, these should be definitely be there in your list. Of course, J advance is subject to your qualification in J main which I don't doubt at all. And maybe you can try for SRM or these local universities which probably you will never join. But just register yourself so that you get exposed to that environment of taking these exams. I personally wrote more than 20. And I actually told you the story. I had to travel all over India to write these exams. In the beginning itself, my dad made all the railway reservations. Four years I was just traveling, traveling, traveling. In my time, there used to be AI Tripoli, there used to be JEE, JEE used to be JEE problems. And then main JEE, main JEE is the advance JEE that you have. Then I wrote West Bengal JEE, I wrote Jharkhand JEE, I wrote Bihar JEE, I wrote Bits, Birla, BIT Meshra exam, that time BIT Meshra used to have a separate exam. Then I wrote CET also, that time CET was open category. So many exams I wrote. Very good, awesome. Very good, Pranav. Aditya, check your working. I think the figures are not correct. Let's check your working. Correct, Kinshukh? Kinshukh, just check the sign of your Y coefficient. Is it the same or the negative or that? Now Kinshukh will like OOF and then he will correct it. So Gayathri, Gayathri you are getting some ugly figures for coefficient of Y. Okay, let's discuss this guys. I think most of you would have definitely attempted it. Okay, Aryan. See, let the circle that I'm looking for, let the circle which I'm looking for, has this equation x square plus y square plus 2 gx plus 2 fy plus c equal to 0. Okay, this circle is basically passing through 4 comma minus 1. And its center lies on x equal to 2 line. Now center lying on x equal to 2 line, what does it indicate? It indicates its center has to be 2 comma something. Because x equal to 2, center has to have abscissa as 2. Let's say the Y coordinate is, normally we know the Y coordinate is minus f. As for this general form of the equation, coordinates of the center is minus g minus f. So you know for sure that g minus g is 2. So g is equal to minus 2. Okay, so this everybody would have bought it. It's not an issue. Second thing is you know that this point 4 comma minus 1 satisfies this. So if you put 4 comma minus 1 on this, you'll get 16 plus 1, plus 8x minus 2y plus c equal to 0. Correct? So this gives you 17, oh sorry, sorry, sorry, sorry. 8g and, yeah, sorry, 8g and minus 2f. So you know your g is minus 2. So g is minus 2 means this is minus 16. So minus 2f plus c equal to 0. So one equation you end up getting is 1 minus 2f plus c is equal to 0. So let's say I call this equation number 2. Let's say this is your 1. So these two equations are very important for us. I hope I have not made any silly mistake anywhere. Do let me know in case anything has gone wrong anywhere in the calculation part only. Now this circle is cut by another circle. Let me make that in white orthogonally. Okay, it is cutting each other orthogonally. So we know that condition between the g1, g2, f1, f2 and c1, c2. So this circle, this circle has g as 5 by 2. So 2g into g1. g1 will be 5 by 2 plus 2f and the f here will be 7 by 2. So 2g1, g2, 2f1, f2 is equal to c1 plus c2. That is minus 4 plus c. Okay, so from here we will get another equation that is 5g plus 7f minus c is equal to minus 4. So this is our third equation. Now we already know the g value. We already know the g value. g value is minus 2. So this will become a minus 10. So this will become a 6. Now add these two. If you add these two, you will end up getting 1 plus 5f is equal to 6. So 5f is equal to 5. So f comes out to be 1. That means 7 minus c is equal to 6. So c comes out to be 1 again. Now we know everything about the given unknowns. So our circle equation required will be x square plus y square plus 2gx. Plus 2gx is minus 4x. Plus 2f y, 2f y will be 2y. Plus 1 equal to 0. So this is your desired answer. This is your desired answer and let me check how many of you actually got it. Ruchitta, you are getting a point circle. Is it a point circle? I don't think so. Aryan, ugly figures, I3, ugly figures. I think Kinshukh now got it correct. No, Kinshukh also made a sign mistake here. Pranam Thirumavadi got it correct. Well done. Yeah, I mean concept wise there is no problem at all. I can understand that. But where you would have probably got wrong is simplification or playing with these three equations or solving these three equations simultaneously. Okay. So with this we conclude the concept of angle of intersection of two circles. And now I'm going to move towards the concept of pole and polar. Very interesting concept. However it is, it will not be asked directly in any competitive exam, but you should be knowing the concept per se. So what is the concept of pole and polar? So pole is basically a point and polar is basically a line. Okay. Now these two, this point and this line are actually linked to each other. How? Let me show the linkage actually. So let's say this is a circle. Okay. Any circle? Okay. If you take a point, let's say anywhere, let's say I take a point here. Let's say this is my point X1, Y1. Okay. Now through this particular point, let me draw all possible sequence to this circle. But for the sake of simplicity, I'll just draw two of them. Whenever I want to draw a line circle, let's say I make one of this is like this. And I make, by the way, this is passing through it. So you can just extend it to the other side. And the other one, I will make it like this. Okay. Just for the purpose of making the diagram, you know, slightly good to look at. So what I'm doing, there is a circle. There is a point X1, Y1. To that point X1, Y1, I'm drawing several sequence. So there can be infinitely so many seconds that you can draw to this circle, isn't it? Now what do you do? The points where these secant cut the circle start drawing tangents to the circle at that point. Okay. So here also I start drawing tangents to the circle at that point. Okay. So I've just shown you two seconds. But how do you draw, let's say millions and trillions of seconds? You'll realize that their meeting point will actually lie on a line. Let me draw that line. Their meeting point will basically lie on a line. It looks like as if it is coinciding. Let me just, yeah. Let me just make it slightly this side. Manual drawing I can change easily. Okay. So let's say it is a situation like this. These two are meeting here. Yeah. Let me just read all the tangents. So these two are meeting here just for the purpose of drawing it. And these two are meeting. Okay. So this line is basically called the polar for this pole. So this is a pole. And this is the polar, right? So definition wise, what is a polar? Polar is nothing but it is the point of intersection of the tangents drawn at the meeting point of the secant drawn from the pole to the circle. Okay. So let me illustrate this process to you on GeoGibra. Okay. So on GeoGibra, I'll make this process. Let me just cancel this off and open this once again. See all of you, just let's take a standard case of a circle. X square plus Y square is equal to nine. That's it. Okay. Now what I do, I choose a point over here. Let's say I choose this point. Okay. Now through this point, I'm going to draw several secants to this particular circle. So let's say I make one like this. Okay. Okay. I make another one like this. I make another one like this. Just making three of them. Doesn't matter whether I make three or thirty or three hundred. That line is going to be the same. Now what I'm going to do at the point of intersection, let's say this point, this point. Let's say this point, this point. Let's say this point, this point. By the way, the one which is in between will go to a large extent. So you'll not see the point of intersection there. So what I'm going to do, I'm going to now draw tangents to the circle at those points of intersection. Tangents here to this circle, tangent here to this circle, tangent here to this circle. Tangent here to this circle. Okay. Tangent here to this circle. As I told you, you'll go very far away. You will not be able to see where it is intersecting. But let me just zoom in a bit. Yeah. Now it has all become kitschy. But what I'm trying to say is just focus on these points. Focus on this point. So here, basically you had drawn a tangent at this point and a tangent at this point, which was meeting at this location. Am I right? Meeting at tangent at this point, tangent at this point. You're meeting at this location. Okay. I hope I have drawn correct. Oh, sorry. This is not this location. Yeah, it is this location, I believe. Yes or no? Yeah. This location, this location and this location. Okay. So I'm just naming it. Focus on a, b and c. Okay. If you connect them, you'll realize that they are all falling on the same line. And this red line that you are seeing on your screen right now. This red line is called the polar for this blue point A, which is your pole. Understood. Does it make sense? Does it make sense? Any question, any concern with respect to that? Okay. Now we will prove it. We'll prove it. Don't worry. So now many people ask me, sir, could this point have been on the circle? Could this point have been within the circle? Now, let me tell you if the point was on the circle, if the point was on the circle, then for that pole, the polar would have been the tangent drawn to that point. We'll see that. If the point would have been inside the circle, again, we'll get a similar kind of a story. We'll see that in another diagram that I will be drawing just now. So let me first draw it on the tool here, on the IPO tool over here. So now I'm going to take another case where the circle is like this. Okay. Let's say I take the circle. Okay. Now, this time I'm taking this X1, Y1 inside the circle. Let's say I take it over here. Okay. And the same story I will repeat. I will draw tangents. I will draw, sorry, I will draw chords passing through this particular point. So let me just draw two of them. I will not draw a lot of them. Let's say this. Okay. And let's say something like this. Okay. So I can draw millions of chords passing through X1, Y1. Now, at the point of intersection, at the point of intersection, if you start again drawing tangents, let's say like this. Okay. Again, I'm just, I'm just drawing a simple situation over here. Not a very ugly one. And you start joining their point of intersection. You would realize this will lie on a line. This will lie on a line. All the points. So I've just shown you two, but if you draw any other chord also and make tangents at the extremities of that chord, they will also meet on the very same line. So this will become the polar for this pole. Is it clear? If the point is on the circle, if the point is on the circle, in that case, the polar will become the tangent drawn to that point on the circle. Let us try to figure that situation also, but now we'll go into that in the equation part of it. So if I know the circle equation, and if I know the location of the pole, can we find out the equation of the polar? This is something which I would like you to first write out. Yes. If the, if the point, if the pole happens to be the center of the circle, then what will happen? The polar will actually go to infinity. Because from a center, if you draw any chord and tangents drawn at the end of the chord, they will never meet. So we'll say the meeting point of all those set of parallel chords will be at infinity. So in that case, the polar will go to infinity. So that's a very extreme case of it. Aditya. Okay. Pranav, I actually explained that probably you didn't hear. If the pole is on the circle, if the pole is on the circle, then the tangent drawn to that point itself is the polar. Do one thing. First figure out what is the equation of the polar. A lot of things will come up. So my question to everybody is plain and simple. Let us say this is my circle x square plus y square is equal to a square. This is my pole x1, y1. Give me the equation of the polar. Polar, polar, whatever you want to call it. Just a pronunciation. What is the equation of the polar? How will you find it out? How will you find it out? Think, think, think. You already know all the concepts. So basically what are you looking? You're looking for the locus of the meeting point of the tangents drawn at the extremities of the chord which has been made through x1, y1. I hope I'm not using a long statement over it. Again, I'll repeat. What am I doing? I'm finding the locus of locus means a path traced by h, k or the path traced by the meeting point of the tangents drawn from the extremities of the chord passing through the pole. Think this to be in a reverse direction. Then you'll be able to, yes, chord of contact exactly or some kinshuk. That's the whole point. The intersection will always be lying on the very same line. That is what we're trying to figure out through the locus. Locus will tell you that it is a line or something else. Yes. See, let's try to figure this out. Have you all done? I mean, anybody who is done with the equation of the polar, if yes, say done. It's just two lines, my dear. Just two lines. Okay, so all of you please pay attention. If I ask you to give me the equation of the COC. Okay. What are the equation of the COC, the chord of contact? You will say the chord of contact drawn from h, k onto the circle equation will be xh plus yk is equal to a square, right? Can I say this COC is satisfied by x1, y1? So x1, y1 will satisfy COC. Correct. That means x1, h, y1, k is equal to a square. This is done. My dear friend, you are done with the answer. Now generalize it. Generalize it. Make your h as x. Make your h as x. Make your k as y. And you're done. So xx1 plus yy1 is equal to a square. This itself is the equation of the polar. This itself is the equation of the polar. Now people say, what? Again? This fellow came again. See, when I was discussing t equal to zero to you, I already informed you that t equal to zero will stand for several types of equation. One is the tangent drawn if x1, y1 is on the circle. Second is it will be the chord of contact if your x1, y1 is outside. So now this is something very important. All of you please realize this, that this polar that you have actually got, it is actually the chord of contact when you connect this point to the extremity. So this actually is the chord of contact also. And it is actually polar also. So if the point is outside the circle, t equal to zero will represent the chord of contact undoubtedly. No doubt about that. But that same chord of contact will also be the polar or polar, whatever you want. Are you getting my point? So the same formula is behaving in two ways. In fact, in three ways, I should say. Number one, it is behaving like the equation of a tangent had this point x1, y1 been on the circle. And that is precisely why I said that your polar will be your tangent drawn to the pole if the pole lies on the circle. Got it now? This equation supports it. If this x1, y1 lands on the circle, this polar will actually be tangent. You can also see, observe that if this starts moving towards the circle, this polar will also start moving towards that point. Second thing is the same equation behaves as the chord of contact also. So from this point, if you draw a chord of contact, the chord of contact will also be the same. Tangent chord of contact polar. They all have t equal to zero. But here you have to see the t is formulated by using what x1, y1? Is it on the circle, outside the circle, within the circle? If it is within, it is completely a polar. If it is outside, polar and chord of contact are same. If it is on, it is the tangent. Polar and tangent will be the same. What's the point? So when the point is within the circle, t equal to zero is completely polar. When the point is on the circle, t equal to zero is polar also. It's tangent also. And when point, pole is outside the circle, t equal to zero is polar also chord of contact also. Liklu, if you want, I will write it down for you. Arjun has a question. The tangents on the pole touch the circle at the points of intersection of the polar. Exactly. So Arjun is saying sir, if the polar cuts it here, then will I say the tangent drawn from there will basically be touching the circle at the intersection of the circle and the polar? Yes. Is there a way to show geometrically, COC is polar? Yes. There is a tool in GeoGebra which helps us to show that, thank you to the GeoGebra community that some enthusiast, coordinate geometry enthusiast has shown that on the tool. So I'll show that to you. Don't worry. I'll show that. Kullaw symptom. Yes. Take off. Enjoy the things over here. Very, very interesting things I am going to show it over here. x square plus y square equal to nine. Okay. Now let's choose a point. Let's choose a point. Let's choose this point over here. Minus four comma minus three probably. Okay. Now there is a button on your GeoGebra which will help you draw the polar for that pole. Which button is it? Let me show you. Click on this perpendicular. Go down. Third last option. Polar. Okay. Click on that. It will ask you to choose the pole first and the circle or the conic. If you see this becomes your polar. So this is, I'm sorry. This is your pole and this is your polar. Okay. Now somebody wanted to know, I think Arjun wanted to know that if I draw tangents from the pole to the circle, will it hit at these two positions? A, B. Let me show that as well. Actions are louder than the words. So there is a tool in the GeoGebra which helps us to draw pair of tangents. I think it is this tool. No. This tool. Yeah. Tangents. Click on that. Click on the point. Click on the circle. There you go. There you go. Awesome. Isn't it? So guys, what did you learn? This is a clear cut indication that COC, COC and polar, they are just two names of the same equation for this point X1, Y1. Clear. Clear. Clear. Clear. Clear. Clear. Okay. I will tell you more interesting things about it little later on. Why didn't we do three classes ago? See I normally see the level of understanding. Then I choose my topics. If I think the level of understanding is not that great. I delayed a little bit. But I definitely don't miss it. Is this fine? Okay. Now let me choose a point inside and also show you the, oh sorry, let me just switch this off Now let me take a point inside, let's take a point here at 1,1 why you people are back delete delete delete Now again if you want to find out the polar for this pole with respect to this circle This is the way to pronounce it, many people pronounce it wrong Polar of the pole with respect to the circle there is no fixed polar for a pole because if the circle changes the polar will also change similarly we say pole of the polar with respect to that circle that with respect to the circle you should not miss out okay so how would you say let me just first draw the polar for this choose the point, choose the curve this is the polar okay so when I say this line is the polar for B I would say with respect to this circle similarly this point is the pole for this polar with respect to the circle are you getting my point is it clear so now let me just go to the the screen and write few things which definitely must be noted down and kept in your mind important observations are noted down the following point number one number one we read this as x1 y1 is the pole of this polar with respect to the circle okay so it's very important that you read this as this okay so this is your pole this is the pole this is the polar but with respect to this circle with respect to this circle okay second important thing that I already told you but I'm just writing it down for you if x1 y1 is outside the circle okay is outside the circle okay then xx1 plus yy1 equal to a square will represent two things one is the polar of this pole with respect to the circle and other is the chord of contact same equation behaving as two things so they will be the same straight line okay if x1 y1 is on the circle then the very same equation will behave as the polar also and tangent at that point okay so tangent itself will be the polar if this point is inside the circle then this will purely represent a polar polar of the pole with respect to the circle is it clear everybody is it clear or not clear okay let me tell you few more things before we jump on a problem in case we have more time we'll take few more questions so if you use t squared equal to ss1 when x1 is inside the circle would we get anything similar to see one small thing I would like to highlight here when you're using it t squared equal to ss1 s1 will be negative in that sense right so what will come out from there do one thing aditya this is a very interesting observation that you should be you know trying to do some research on take the point inside okay you know what will happen on or outside that is fine take the point inside and try to figure what is t squared equal to ss1 boiling down to because from here the analysis may not be very very sharp because I'm just able to say that t squared is equal to s into a negative quantity that is what I can say and what type of equation will come out from there that is something which we need to look into you take a generic form only don't take a special form take a generic form see what it is some conic it is some conic coming out okay anyways you've got a you've got an oval shaped section okay we'll figure that out because that is something which even I have not figured it out because it is not useful it doesn't nobody ask that type of concept in the questions also okay now let's analyze something very important very very important so this is point number two point number three point number three if if x1, y1 and x2, y2 are such points are such points such that let me write our points such that our points such that x1 lies on the polar of this and vice versa okay so basically let's say I call this point p and I call this point q right so p lies on the polar of q and q lies on the polar of p so p and q lie on each other's polar okay then then p and q are called conjugate points they are called they are called conjugate points is this fine now let me show this demonstration to you let me show this demonstration to you see all of you please have your attention on the screen right now now you can see that b is a point 1 comma 1 okay and this is the polar of it on now this polar I will choose a point let me choose a point on the polar here okay now with respect to a if I draw a pole for this particular circle let's say choosing a as my pole I make a polar with respect to this circle you would realize that that polar will actually pass through the b point so this is a situation where a is lying on the polar of b with respect to this circle let me write it down here a is lying on the polar of b and b is lying on the polar of a okay then these two points will be called as the conjugate points they will be called as the let me use a different thing they will be called as the conjugate points is it clear any questions here anybody so there are two points x1, y1, x2, y2 says that both lie on each other's polar then those two points will be called the conjugate points clear any questions sir could these two yeah these two polars can intersect somewhere they may be parallel also when they're really parallel I'll tell you that is a special location wait so any point on these will have polar through b any point on b's polar will have polar through a yes so there will be several set of you can say there will be several pairs of conjugate points possible so let's say if I choose another point or for that matter I shift this point let me choose that all of you please pay attention if I shift this point see the polar will also get rotated about b only do you see that the polar of a is getting rotated about b so in this case whatever point you are shifting your a on those pairs will be conjugates conjugate so every point every point on the polar of b is its conjugate got the point is it clear so wherever I took take this point you can see the polar of that point will always pass through b are you getting my point is it clear could you repeat this point see if I take a point a here I get a polar of a passing through b so a and b are conjugate pairs or conjugate points if I took it here still the polar of a is passing through b only that means a and b are still conjugate points but now a point position has changed so now whatever point is it lying on that point and this point will be conjugate points of each other so every point on every point on the polar of b will be basically a conjugate point of b or they will be conjugate points so whether you take this point this point this point this point they will all be the conjugate of this guy b that is what I am trying to okay now I know many of you would be doing lot of experimentation same will be true for b also okay Hariharan wants me to move b let's move b if you move your b okay b is actually let me move b on the actually b has let me do the other verum because I am moving the polar also by doing that so that is getting changed so let me do another figure just once again Hariharan I will show you the let me make the circle again okay let me take a point this time outside the circle okay let's say a now I am making the polar of a with respect to the circle okay now take any point b on this circle let's say I take a point b here or somewhere over here okay and I now make the polar of b with respect to the circle it will pass through a as we have already seen that now Hariharan as per your requirement I will move b you will see that the polar will still pass through a only it will not change clear Hariharan when to the two polars be parallel that's actually a question I wanted to ask you but ask the time but will I tell you that theory also that's a very good observation Arjun now all of you please pay attention the fourth point is the fourth point is the fourth point is I don't have much time so let's let's discuss it if if l1x plus m1y plus n1 equal to 0 l2x plus m2y plus n2 equal to 0 are such lines are such lines that are lines such that are lines such that the pole of one lies on the other then the two lines are called conjugate lines are called conjugate lines now conjugate lines comes from the very same diagram let's go back again to GeoGebra so as you can see these two lines are conjugate lines because their poles respectively lie on the other but remember if you vary this point even the conjugate line pairs will vary okay so these two points I mean just at the present situation these two lines are called conjugate lines and these two points are called conjugate points that is what we need to what that is what we have discovered so far these two points are called conjugate points okay clear any problem with respect to conjugate points and conjugate lines I'll take a snapshot of this and put it I've seen some questions being asked not in J main though maybe in J advance or something okay so these two are conjugate points and these two are conjugate lines is it fine any questions lines by the way I wrote a line minus take lines okay next important understanding that you will all have number five I'll take some point take some time more so please bear with me if I take five more minutes of your time next important thing that you realize is that if you have a circle okay just hold on your questions for some time okay hold on your questions for some time if you have a circle and there is a center O and let's say this is my point P okay x1 y1 okay and I draw a polar with respect to this point P with respect to the circle and let's say the polar lies here sorry I wanted to make a line two things you will observe one is if you connect your center to P and extend it it's a very obvious observation you would realize that you would realize that number one OP sorry OQ or the line connecting the center to the meeting point of the polar will be perpendicular to the polar so this will be making 90 degree and second point you will observe is that OP into OQ will be always the square of the radius of the circle so let's say this circle is having a radius of r so OP into OQ will always be r square now I will quickly prove it first of all if you know this point is x1 y1 you easily know the equation of this polar that is xx1 plus yy1 is equal to r square so what is the slope of the polar slope of the polar will be negative x1 by y1 correct so this is also the slope of let me write it down let me write it down like this what is the slope of OQ or OP whatever you want to call it what is the slope of OP or OQ slope of OP and OQ can be easily figured out because this is origin and this is x1 y1 so you can easily say y1 by x1 now if you take the product of these two that is m of the polar and m of the OQ line you will end up getting negative x1 by y1 into y1 by x1 cancelling it out and giving you a minus 1 clearly indicating that OQ is perpendicular to the polar is this fine any questions now all of you please pay attention the second point is very important because direct questions have been framed on this OP into OQ is basically r square how that is also very easy to prove only we know is under root of x1 square plus y1 square it is just the distance between origin and that given point and what is OQ OQ is nothing but it is the length of the the distance of the origin from this polar so what is the distance of the origin from the polar so you will put 0 plus 0 minus r square mod by under root of x1 square y1 square you will automatically see that your denominators will get cancelled and you will end up getting mod of negative r square which is actually r square and that proves the second part of the formula before I end the story for today a very important observation if you draw a polar for Q point with respect to the circle somebody was asking this question to me when will the conjugate lines become parallel you would realize that this conjugate line will now become parallel to this so this is basically this is basically polar of P and this is polar of Q so polar of P and polar of Q will now become parallel to each other and in such situations when this is happening we say Q is the reflection of P on the circle so for such conditions we say P is the reflection of Q and vice versa and vice versa with respect to the circle and vice versa with respect to the circle okay now let me show that to you first make a note of whatever I have written so two things one important thing is that if you connect O to P and extend it to the polar of P the line segment OQ or the line OQ will be always perpendicular to the polar that is number one second thing is if you take OP extend it to meet the polar at Q OP into OQ will always be the square of the radius of the circle and P and Q will be called as reflections of each other about the circle that means if you take that circle as a mirror if you take that circle as a mirror then the reflection of P will be Q, reflection of Q will be P vice versa, object and image gets interchanged, we all know that okay so that feature is there in GOG graph let me show that to you it's very interesting observation guys all these things which I am saying is missing from your J ebooks no matter whatever J ebook you are following whatever I have told so far I mean with respect to the reflection and all that is not there in J ebooks okay so let me just show that you know illustration to you okay yes let us choose a point inside here okay let me choose a point inside here 2, 1 now with 2, 1 I would draw a polar this is the polar now what I am going to do from the origin I am going to extend or one more thing important thing in this GOG graph there is a feature called reflect object in circle so third last option third part reflect object in the circle right click that it will ask you for the point and it will ask you to choose the circle if you do that you will automatically get A dash as the reflection okay now I would like to show you that if you had drawn a polar for A dash it would give you a line which is passing through A and parallel to the line which you are seeing on your screen there you go okay so if you see slope here is minus 2 slope here is also minus 2 so they have become parallel okay so somebody was asking when do the conjugate lines become parallel the conjugate lines become parallel when the conjugate points are reflections of each other about the circle okay and another thing that you would all observe from the figure maybe I will use to exhibit that if I connect the center of this circle if I connect the center of this circle to A it will cut the polar at A dash so let me just make that lines from this is it clear and this product this product let me call it this into this will be r square so OA dash into OA will be r square so under this situation you would realize that the conjugate lines will be parallel to each other this is something which I would like you to prove here you have seen the demonstration but what is the proof that if two lines satisfy this or if the two points satisfy this relation the respective polar of both these points are parallel to each other okay so please take this as a proof so I just do one thing I will take a snapshot of this and I will put it as a question for you to solve sorry I am not reading your doubts right now I am just giving the question and then I will explain you your doubts okay so the question here is if OA is into OA dash is equal to r square then prove that the polar of A and polar of A dash are parallel this I am giving you as a homework question it is very easy you will be able to do it now I would take questions from you do you mean the spherical mirror yes okay so please do this and that is all for today all the best for the people who are writing their UT's to