 So hello everyone. Great. Good morning to all of you. Hope you all are doing good in your studies, right? So we were discussing circles, right? So we will take up the exercise seven of circles and we will finish this exercise. Okay. So we were like, after this, we will be like, we are done with this circle chapter. So this is the last exercise. So let's begin this exercise and do the questions. Okay. So here comes the first question. It is saying the circles x square plus y square plus x plus y equal to zero and x square plus y square plus x minus y equal to zero intersect at an angle of. Okay. Two circles are given and the question is asking their angle of intersection, right? So let's first check it out whether these two circles are orthogonal or not. This is the best way we can start with. So let me write the equations. So here is our first circle. This is given as x square plus y square plus x plus y equal to zero, right? And our second circle is x square plus y square plus x minus y equal to zero. So what I am saying, I will start with checking the orthogonality of these two circles. Okay. So in circle one, if you say, in circle one, if you say, what is our G1? Like here, if you see our G1 is coming out to be that is one by two, right? And what will be our F1? Our F1 will also be one by two. And in this second circle, if you see our G2 will be one by two and our F2 will be minus one by two, right? And for orthogonality, we know that there's two G1, G2, right? This condition should satisfy for the two circles to be orthogonal, two G1, G2 plus two F1, F2. This must be equal to C1 plus C2. What is C1, C2? C1, C2 are the constant terms in the equation of the circle. So let's put the value of G1, G2 and F1, F2. What we get? We, to get two into G1 is half, okay? G2 is also half plus two into F1, F1 is half and F2 is minus half, okay? And what is C in this equation? It is zero. And C1 is zero and C2 is also zero. So it will be zero plus zero, okay? So these two, two gets canceled out. These two, these two gets canceled out. So it will be basically half minus half is equal to zero. So zero is coming out to be equal to zero, right? It means it is satisfying this equation. It is satisfying this condition, this condition of orthogonality, right? This is the condition of orthogonality. Orthogonality. So since this, this is getting satisfied for both these circles, we can say that both these circles are orthogonal, okay? Means they are intersecting at an angle of pi by two. So this option D is correct, okay? Now let's take the next question. Here it is saying if the circles of same radius A and centers this and this could orthogonally then A equals two. Now here what has been given? The centers of two circles are given. The first center is two comma three. This is our center of a circle, okay? That is two comma three and our center of second circle is given as five comma six, okay? And both are having radius A, okay? And both the circles are cutting each other orthogonally. Then question is asking to find the value of A, okay? So how should we proceed? We should proceed with writing the equation of the circle itself. So let me write the equation of first circle. It will be basically x minus two whole square, right? Two and three is the center of this circle. So x minus two whole square plus y minus three whole square. And that should be equal to radius squared. And what is the radius? Radius is given to be A. So it should be equal to A squared, right? Let me write the equation of second circle in this side. So what will be the equation of second circle? It will be x minus five whole square, right? x minus five whole square plus y minus six whole square and that should also be equal to A squared. So this is the equation of our second circle. This is the equation of our first circle. Now we know for these two circles to be orthogonal what we discussed in the last problem that two G1, G2 must be equal to C1 plus C2, okay? So let's expand this first, okay? So what we will get? The equation of circle one, we are expanding, right? So this will be x square plus y square from here, then minus of four x, right? Minus of four x, it will become minus of six y and plus four from here and plus nine means 13. So this will be 13 and take this A squared in the center. So this 13 minus A squared is equals to zero. This is our S1. And what will be our S2? Our S2 will be basically x square plus y square, then minus 10 x, okay? From here we get minus 12 y, then plus 25 plus 36, 46, 56, 61. So this will be 61 and minus A squared is where I'm taking it to left-hand side. So this will be equal to zero, okay? So this is our S1, means equation of first circle. This is our equation of second circle. So we can easily find the value of G1, G2 and F1, F2. So what will be the value of G1 from here, if you see? It will be basically minus two, right? And what will be the value of, sorry, what will be the value of F1? The value of F1 will be minus three, okay? And what will be G2 from this equation? It will be minus five and what will be F2? F2 will be minus six. So we know the condition of orthogonality and this is our C1, right? So this is our C1 and this thing, this constant thing will be our C2. Okay, don't get confused with this C1 and C2. Okay, let me write it as C1 dash and C1 dash. Like these are representing centers. These are representing centers of the circles. So don't get confused with this. So now I'm writing this two G1, G2, this two G1, G2 plus two F1, F2 must be equal to C1 plus C2, right? So put the value. I will get two into minus two, G1 minus two, G2 is minus five. Then two times F1, F1 is minus three. What is our F2? F2 is minus six and this must be equal to C1. This is 13 minus A square, right? And what is our C2? This is 61 minus of A square, okay? Solve it four, two to four, four into five will be 20. So this will be 20 plus 66, 36. And in the right-hand side, we have 61 plus 13, 71, 74, 74 minus two A square. So from here, we get two A square is equal to 74 minus this 56. So 66, 76, 18. So from here, we get A square is equals to nine, right? If A square is equal to nine, A square is nine, so we get A as plus minus three. Okay, this will be our answer. It is asking the value of A one bit. So the value of A will be plus minus. So it's given three here in the option. So let's check this option as correct. Now, see the next question, question number three. We are saying if the circles, this x square plus y square plus two x plus two ky plus six equal to zero. And one more circle is given here, x square plus y square plus two ky plus k equal to zero. Intersect orthogonally. Same thing. We are given, we are provided with the two circles. And the question is saying that these circles are intersecting orthogonally, then we have to find the value of k. So the next question we will apply here also. Okay. So let me write the equation of the first circle. It is given as x square plus y square plus two x plus two ky plus six is equals to zero. Okay. And what is our second circle? Second circle is x square plus y square plus two ky, right? Two ky plus k is equals to zero. Okay. So what is the value of G1 here? It will be one only. And what will be the value of F1 here? It will be k, right? And what is our G2? G2 here, the x term is missing. So G2 is zero. And what is F2? F2 is k. Okay. And let's write the value of C1 also. So C1 from here, what we get? Six. And C2 here, we get k. Okay. So apply the condition that is two G1, G2. Two G1, G2 plus two F1, F2 must be equal to C1 plus C2. So two G1 is one. What is G2? G2 is zero. Then two F1, F1 is k. What is F2? F2 is also k. Okay. This is as per the question only. So two F1 into F2 and that must be equal to C1 plus C2. That is six plus six plus k, right? So it will be forming. This thing will go off. Okay. So it will be basically a quadratic in k. That is this two k square minus k minus six is equals to zero. Right. So six and two, two, twelve. So we can split it as two k square. Okay. Minus four k plus three k minus six equal to zero. So take two k common here. It will be a k minus two. Take three common k minus two equal to zero. So k minus two into two k plus three is equals to zero. So what we got here, either k is equal to two, right? Or k is equals to minus three by two. Right. So this is the, this will be the value of k. These two values two and two are minus three by two. So option is correct. Okay. Same condition we have applied in all the, these three questions like condition of orthogonality. This two g1 plus g2 is equals to, means two g1 g2 plus two F1 F2 equals to C1 plus C2. Okay. Now let's see this fourth question. It is saying if a circle passes through the point this, okay. And curts the circle X square plus Y square equal to four orthogonally, then the locus of its center is. Okay. Like one circle is given here, X square plus Y square is equal to four. And the question is asking to find the locus of the center of that circle, which cuts this circle orthogonally and that circle passes through this point A comma B also. Okay. So let's do in this way like we are given one circle, X square plus Y square minus four equal to zero. Okay. And like, let me assume, let me assume, let our second circle means let the required circle, let the required circle be of this equation. The equation of the required circle being X minus H whole square, right? X minus H whole square plus Y minus K whole square is equal to R square. Since we have to find the locus of this required circle, this. Hence I have considered this equation. And finally what we will do, we will replace this H and K by X and Y. Okay. Now this circle, this circle is passing through, it passes through. It passes through the given point. Let me assume that point as B whose coordinates are A comma B. Okay. So this circle, this point must satisfy this equation, right? So let's put this. So what we will get? What we will get? Or we can do one more thing, we can expand this equation, right? We can first expand this equation, then we will put the value. So let me expand it and write it. So it will be basically X square. Okay. Plus Y square from here, then I will get minus two HX, okay? From minus two KY. And from here I will get H square, okay? H square plus K square and minus R square. Okay. This is equals to zero. Now what I will do? I will put this point B in this equation. Okay. So putting this in this equation, what I will get? This will be basically A square plus B square and then minus two AH, okay? Then minus two BK, right? Plus H square plus K square minus R square is equals to zero. Right? This is what we got. And one more information is given like these two circles means this S1 and S2. Both are orthogonal, right? Both are orthogonal. So one more condition will be satisfied by these two circles. Okay. What is that condition? This is two G1, G2, right? Plus two F1, F2 must be equal to our C1 plus C2. So here you observe in the first circle, G1 is equals to zero and F1 is also equals to zero. So this thing in the left hand side will finally become zero. I don't know what is G1, G2 and F2, but since G1 and F1 are zero here, it will finally become zero only. So this will be equal to C1. What is C1 here? It is minus four. And what is C2? You see C2. What is C2 in this equation? This is our S2. This is our S2. So what will be our C2? This will be basically h square plus k square minus R square. h square plus k square minus of R square, right? So from here we got the value of this whole thing. h square plus k square minus R square is equal to four. Now what I will do? I will put this value in this equation. So what I get? I will get this thing a square plus b square, okay? And this whole thing is equals to four, right? So plus four is equals to this thing, two a h plus two b k, right? Now replace this h and k by x and y. So h and k will be replaced by x and y and we finally get our equation of the locus of center, right? So this will be basically a square plus b square plus four is equals to two a x plus two b. This is our required locus. So two a x plus two b y is equal to a square plus b square plus four. So a square, okay, option b, option b is matching, right? This two a x plus two b y is equal to a square plus b square plus four. So hope it is clear to everyone. We are done with this. Let me move to the next question. So yeah, this is our question number five. It is saying the locus of the center of the circle which cuts orthogonally this circle x square plus y square minus two x, 20 x plus four equal to zero and which touches x is equals to two, okay? So it's like it is given. One circle is given here. This equation is x square plus y square minus 20 x plus four is equals to zero, right? And we have to find the locus of the center of that circle which cut this circle orthogonally and touches this line. So before finding locus, let me like assume, let us to let me assume the equation of that circle first, okay? And here also I will take in the form of center and radius since we have to finally find the locus of the center of circle. So let s2 is our x minus h whole square, right? Plus y minus k whole square is equal to r square, right? And we will also need the value of g1, g2 and f1, f2 so better to expand this one, this equation. After expanding what I will get x square plus y square, okay? Minus two hx from this first term like then minus two k minus two ky, okay? Then h square plus k square minus r square, okay? This is equal to zero. So I am assuming the second circle to be this one. So since both these circles are like orthogonal, so that condition will be satisfied, the condition of orthogonality. So what is g1 here? g1 is basically minus n and what is f1 here? f1 is zero, okay? Now see this value of g2. What will be g2? It will be minus of h, right? Minus of h and our f2 will be minus of k. So this two g1, g2 plus two f1, f2 equal to c1 plus c2. So this g1 is minus 10. What is our g2? g2 is minus of h and this will be two into f1. f1 is zero so anyhow this is going to be zero only. Okay, anyhow I have written that. And what is our c1? That is four and what is our c2? c2 is this whole thing, right? So this will be four plus this thing. h square plus k square minus r square, okay? Now this is going to be zero. And what we got here? Minus means 20, 20 h, right? So we got 20 h is equals to four plus h square plus k square minus r square, right? So this is one equation which we got from satisfying the condition of orthogonality. Now one more information is given here that this circle s2 touches x is equals to two. So suppose this is our like touches x comma two, okay? So x equal to two will be a tangent basically. So if you see, let me draw this, this tangent is x is equals to two means x minus two equals to zero. And what is the center of this circle? This s2, this is h comma k basically, right? The center of this circle. So what I will do? I will drop a perpendicular from the center on this tangent, right? And this will be, and I will equate this distance to radius of this circle. So hope we will get something. Let's try that. So suppose I'm taking this as p. So what will be our p? p will be basically one into h that is h then zero minus of two, right? Minus of two, this will be more. And under root of a square plus b square that is one. And that must be equal to radius. What is radius of this circle? This is r. This is r basically. r is the radius of this circle, no? So from here we got h minus two is equals to, right? h minus two is equals to r. Okay. Now what we can do? What we can do here? This h minus two equals to r. We can do like, we can substitute the value of r in this equation. Okay. If you see, I'm substituting the value of r here. So we get 20 h is equals to this thing, four plus h square plus k square. And what is r square? From here, if you see r square is equals to h square plus four and h square plus four minus of four h, right? So write it minus of h square minus of four plus of four h. Wow. This four and minus four is getting cancelled out. This h square minus h square is getting cancelled out. So finally we got this thing. 16 h is equals to k square. Now replace h by x and this k by y. Now we got this thing. Y square is equal to 16 x. So this will be the required numbers. Y square is equal to 16 x. Y square is equal to 16 x. So this option says correct. Okay. Take this next question. Question number six, right? The equation of a circle which calls the three circles. This, this and this orthogonally. Find the equation of a circle which is orthogonal to all these three given circles, which is orthogonal to all these given circles. Okay. So actually if you see three circles are given here, no? And we have to find the equation of circle which is orthogonal to all these. So basically the radical center of these three circles, the radical center of these three circles will be the center of the required circle. Is it okay? I think you have learned that this concept in the while, while you were studying this radical axis and coaxial system of circles. Okay. And then it will be like, I will be finding the radical center of these three circles. Okay. So this, let me write first the equation of the circle. So our first circle is x square, okay, plus y square, then minus three x minus six y plus 14 is equals to zero. Okay. The second circle, is it okay? No, see it x square plus y square minus three x minus x. Then write the second circle, equation of second circle, it will be x square plus y square minus of x minus of four y plus eight is equals to zero. And what is our third circle? Third circle will be, means given as x square plus y square plus minus six y plus nine is equals to zero. So these three circles are given in our question. Now, what I will do, I will find the radical axis of this S1 and S2. Radical axis of S1 and S2. Okay. This will be, how can I get this radical axis of S1 minus S1 and S2? It will be basically S1 minus S2 is equals to zero. When subtracting both these circles, I will get the equation of radical axis of S1 and S2. And what will be radical axis of this S2 and S3? What will be radical axis of S2 and S3? It will be basically S2 minus S3 is equals to zero. And the intersection point of these two radical axis will give me the radical center. So let me write it as radical axis one and let me write it as a radical axis two. Okay. So find this radical axis one. Radical axis one will be S1 minus S2. So this x square y square terms will be cancelled out. So finally I will get this minus 3x minus of minus x means plus x. So I will get minus of 2x and minus 3x plus x minus of 2x minus 6y plus 4y. So it will be basically minus 2y and plus 14 minus 8 will be 6 is equals to zero. Okay. So from here I get x plus y, x plus y minus 3 equals to zero. Okay. Let me assume it as equation one. And what will be the radical axis second? Second radical axis will be basically S2 minus S3. So it will be minus of 3x then plus of 2y plus of 2y and 8 minus 9 that will be minus 1. Okay. Minus 1 equals to zero. Okay. Then you can write it simply as 3x minus 2y plus 1 equals to zero. Let me assume the equation two. This solving RA1 and this radical axis two will give me the radical center. So let me multiply it by two in the first equation. We get 2x plus 2y, 2x plus 2y minus 6 equals to zero. And here we are having 3x minus 2y plus 1 equals to zero. So add it. So we get 5x minus 5 is equals to zero. Means x we are getting as 1. We are getting x as 1. And if x is 1, what will be our y? X is 1 means y will be 2. Y will be 2. So what we got? We got finally radical center, radical center of these three circles. Okay. Of these three circles. So we got radical center. Let me write it as radical center of these three circles are 1 comma 2. Now this will be the center of the required circle. This will be. I'm writing also this will be. This will be center of center of required circle. Required circle. Okay. So what will be our required circle? What will be the equation of our required circle? So a required circle will be basically x minus 1 whole square. Okay. x minus 1 whole square plus y minus 2 whole square. And is equal to R square. Now, what will be this R means what will be a radius of this center? What will be radius of this circle? It will be basically the distance means length of tangent from this center to any of the circles like three circles are given here. You can take any of the circles and you can find the length of the tangent from this center point. It will be the radius of this circle. It will be the radius of this circle. So basically R will be equal to under root of S1. Okay. Or under root of S2. Or under root of S2. Or it will be under root of S3. Means point which is our radical center 1 comma 2. Its power of point, you can take any of the circles that will give you the radius of this circle. Okay. So I am putting it on equation 1. Like on S1. So what we got? 1 square means 1 square plus what is our y coordinate 2. So this will be 2 square minus 3 into 1 minus 6 into 2 and plus 14. This will be the length of tangent. Or in this case particular in this case it will be length of the radius. So it will be 1 plus 3. 1 plus 4. 5 minus 3 means 5 minus 3 minus of 12 plus 14. Okay. So 14, 5, 19. 15 minus 4 means under root of 4 that is equal to. Okay. So finally what we got? Finally we got the circle as x minus 1 pole square plus y minus 2 pole square is equal to r square. What is our r is 2? That is 2 square. So this will be the answer. This is the circle which are cutting these 3 circles orthogonally. So answer is in expanded form. So expanded. x square plus y square from here I get minus of 2x then here I will get minus of minus of 4. And 1 plus 4 that is 5. 1 plus 4 and put this thing here minus 4 equal to 0. So this plus 4 minus 4 will be cancelled out. x square plus y square minus 2x minus of 4y plus 1 equal to 0. So this is our answer. x square plus y square minus 2x minus 4y minus 2x minus 4y plus 1 equal to 0. So I think option A is matching. So this will be our required answer. Now let's take this next question, question number 7. It is saying the equation of radical axis of circles. Okay, 2 circles are given here and we have to find the equation of radical axis. So it is nothing but we can obtain the radical axis by subtracting these 2 circles. Then subtracting the equation of these 2 circles. So basically our required radical axis will be s1 minus s2 is equals to 0. And what is our s1? This thing x square plus y square plus x minus y plus 2. This is our s1 and what is our s2? Okay, s2 we have to first divide by 3. So it will be basically x square plus y square minus 4 by 3x minus of 4 equal to 0. This will be the equation of radical axis. So finally we got radical axis as this x square minus x square will be cancelled out. y square minus y square will be cancelled out. This will be x plus 4 by 3x, right? Then minus of y, no y term here. And plus 2 and minus of, means plus 6. Minus, means plus 2 plus 4 that will be plus 6 equal to 0. Okay, so take L sin 3 that will be basically 3x plus 4x minus 3y plus 18, right? Plus 18 is equals to 0. So this will finally become 7x minus 3y plus 18 is equals to 0. 7x minus 3y plus 18 equal to 0. This is our radical axis of the given circles, right? So option B is correct. Okay, let me take next question. The radical center of circles. Okay, 3 circles are given here and we have to find the radical center. So similar type of question we have done in earlier examples also, right? Nothing to worry. Let's write the equations of the circle. So our first circle is this x square plus y square minus 1 equal to 0. Our second circle is x square plus y square, right? x square plus y square plus 10y plus 24 equal to 0. And our third circle is x square plus y square minus of 8x, right? Then plus 15 equals to 0. Now first find the radical axis of S1 and S2. Radical axis of S1 and S2, okay? Then we will find the radical axis of this S2 and S3. Radical axis of S2 and S3, okay? And we will take the intersection point. Let me write it as radical axis 1. Let me write it as radical axis 2. Then we will find the intersection point, intersection of RA1 and RA2. That will give me the radical center, right? This is our approach. This will be our approach. So what will be radical axis of S1 and S2? It will be S1 minus S2 equal to 0. So this will be basically minus 1 and minus, minus of what? Minus of 10y, right? 10y plus 24 is equals to 0. Am I doing right? Yeah, minus 1, minus of, okay? So finally we will have minus 1, minus 10y and minus of 24 equal to 0. So from here we get this minus 25 equal to 10y. Means why I will be having minus of 5 by 2, right? Why I will be having minus of 5 by 2? And what will be radical axis of this S2 and S3? Okay. It will be basically 10y plus 24 minus of minus 8x and then plus 15 is equals to 0. So from here we get 10y plus 24 plus 8x, okay? And minus 15 equal to 0. So what I get? This thing 8x, okay? 8x plus 10y and plus 24 minus 15. This will be plus 9, no? Yeah, plus 9 equal to 0. So this is our second radical axis. This is our first radical axis. Okay, let's solve it. So our 8x, okay? 8x plus 10, what is y? y is minus 5 by 2 and plus 9 is equals to 0. So this 2, 5 means 8x minus 25 plus 9 equal to 0, minus 25 plus 9. So this will be 16, no? Minus 16. So 8x will be equal to 16. From here we got x is equals to 2 and y we already knew. So minus 5 by 2, this will be basically a radical center, right? This will be our radical center. So x equal to 2, y equal to minus 5 by 2, x equal to 2. 2 will come, no? This 8x minus 25 plus 9. That will be minus 16. I'm going right hand side, it will be plus, okay? So x is 2, y is minus 5 by 2. So option, option D is correct. Now what is given here? If 1 by 2, 1 by 2 is a limiting point of coaxial system of circles containing circle this, then the equation of the radical axis is okay. So basically this limiting point, limiting point is given as, limiting point is given as, let me call it L, okay? So L is given as, coordinates is given as 1 comma 2, okay? And our second circle is, and our second circle, the equation of circle is given as, x square plus y square, right? Plus x minus 5y plus x minus 5y plus 9 equal to 0, okay? And what is it is asking? It is asking to find the equation of radical axis. So basically this limiting point, this is basically what? What is this limiting point? This is actually circle, point circle you can say, point circle. Point circle whose centers are this one, having center, point circle just a center, having center, this 1 comma 2, okay? Having center is having center 1 comma 2 and radius of this point circle will be equal to 0. Radius of this point circle will be 0. So I can write this point circle as, I can write this point circle as this thing, limiting point what is given here. I can write this as, in terms of circle, I can write it as x minus 1 whole square, right? x minus 1 whole square plus y minus 2 whole square is equal to 0 square. Why is 0? Because it is a point circle and its radius is 0, okay? So what will be our s1? Our s1 will be basically x square plus y square, okay? From here we get minus 2x, from here we get minus of 4y, then plus 1 plus 4 that is plus 5 is equals to 0, right? So this will be our equation of this point circle. Now the question is asking to find the equation of radical axis, right? So radical axis we know, radical axis is nothing but s1 minus s2 is equals to 0, okay? So what we will get? We will get this thing, minus 2x minus of 4y plus of 5, right? And minus of this x minus 5y plus 9 is equals to 0, okay? Now open it, it will be basically minus 2x, minus x, minus 3x, minus 4y plus 5y. It means plus of y and plus 5 and minus 9, right? Minus 9 plus 5 will be minus 4, minus 4 equals to 0. Take minus sign common, it will be 3x minus y plus 4 is equals to 0. 3x minus y plus 4 equals to 0. So option b is correct. So this will be our radical axis, right? This will be our radical axis. So I think this question is clear to all, right? So let's move. So let's take the next question that is question number 10, right? So the question is saying the limiting points of the system of circles represented by the equation this, okay? Like equations, system of circles represented by the equation. So equation of a set of circles is given here, which I'm writing here 2 into x square plus y square, okay? Plus lambda x plus 9 by 2 is equal to 0. This is 9, 9 by 2 equal to 0. Now dividing throughout by 2, we get this x square plus y square plus lambda by 2x, right? And plus 9 by 4 equal to 0. Now we have to find the limiting points of the system of circles. This will basically represent a family of circles, like changing the value of lambda, we will get different equations of the circle. Now for finding the limiting points of system of circles, like what does this limiting points generally, what does this limiting point mean? This is what I have said earlier also. This is what you say, point circle, these are point circles with radius 0. These are point circles, right? These are point circles with radius 0 with radius is equal to 0, right? So actually if you see, if you see in this equation, right? What will be g? Like 2g is equals to lambda by 2, right? That means g will be equal to lambda by 4. And what will be f here? f is 0 basically. So f is 0 and our c is 9 by 4. So what will be radius of this circle? What will be radius of this circle? Of this system of circle. So radius of this circle will be basically under root of g square means lambda square by 16 plus f square 0 and minus of c. That is minus c and that must be equal to 0. So from here what we get? We get this thing. So lambda square, okay? Lambda square by 16 and minus 9 by 4. I am squaring it basically is equal to 0. So from here we get 16, no? So this will be basically lambda square minus 36 is equal to 0. So from here we get lambda is equal to plus minus 6. We get lambda is equal to plus minus 6. And what we need to find? We need to find the limiting point, right? So we need to find the center of this set of circles. So center will be basically minus g comma minus a, right? So this thing will be our g is lambda by 4. So our center will be minus lambda by 4. What is f? No y term means f is 0 basically, right? So this will be our center and that will be the limiting point basically. So minus lambda by 4, okay? I am taking this as when lambda equal to plus 6, okay? So when lambda is equal to plus 6, okay? And we will get the limiting point as, limiting point as minus of 6 by 4. That is minus of 3 by 2, right? And when lambda is minus 6, when lambda is minus 6, right? So our limiting point will be minus of minus 6 by 4, right? That is nothing but 3 by 2. So this will be our limiting points, right? Plus 3 by 2 and minus 3 by 2. Plus 3 by 2 and minus 3 by 2, that means this option a is correct, okay? Now let's take this next question, question number 11. Here one of the limiting points of coaxial system of circles containing these circles, okay? So one coaxial system of circles is given here. Those two members are these circles, this x square plus y square minus 4 equal to 0. And the other member of this coaxial system is x square plus y square minus x minus y equal to 0. And the question is asking to find the one of the limiting points, like one of the limiting points. So see here, our first member of coaxial system is this circle, this x square plus y square minus 4 equal to 0. And who is our second member? Our second member is this, x square plus y square minus x minus y equal to 0, okay? So I can write the family of the equations, like family of this coaxial system of circles. How can we write? We can write this family equation as family of circles equation as this thing. S1 plus lambda S2 is equal to 0, okay? Or we can do one more thing. We can write this the same thing as this also. S1 plus lambda, right? S1 minus S2 equal to 0. You can consider anyone, okay? No? So this will be representing the family of this coaxial system of circles. Then what I have to do? I have to do find the limiting points. Then after getting this circle, what I will do? I will liquid the radius of that circle to 0. Then I will be able to find the limiting point. This will be our approach, right? So let's consider this second one. So what I will do? This x square plus y square minus 4, right? And plus lambda times. What will be S1 minus S2, basically? S1 minus S2 will be basically minus 4, minus of x square and y square will be cancelled out. That's why I'm not writing here. So this will be minus x and minus y equal to 0, right? So this will be our S1 minus S2. So not equal to 0. Don't write this equal to 0. So this will be minus 4, right? Is minus 4 plus x and plus y equal to 0. Or we can say in this thing, this x square, x square plus y square, right? Then plus of lambda x, then plus of lambda y, lambda x plus lambda y, then minus 4 lambda, minus 4 lambda and 4 from here. That is minus 4 equal to 0. Now this is the equation of the circle, one of the circle of the family. Now for finding this limiting point, what I will do? I will equate the radius of this circle to 0. So here if you see, our 2g is equals to lambda, right? And our 2f is equals to same lambda. So our g will be basically lambda by 2. And our f will be also lambda by 2. And our c is minus 4 lambda minus 4. Okay. So what will be radius? So what will be radius? Radius will be under root of g square plus f square minus. And that will be equal to 0 for limiting points, right? For finding the limiting points. So g square means this lambda square by 4 plus f square also lambda square by 4. And minus of c, that will be minus of c means minus of this whole thing. This whole thing will be then plus 4 plus 4 lambda, right? 4 lambda is equals to 0. So this will be basically lambda square by 2, right? Lambda square by 2 plus 4 lambda plus 4 equal to 0. Or we can say lambda square plus 8 lambda, right? Plus 8 lambda plus 8 equal to 0. So this is coming out to be a quadratic in lambda, right? And from here if you see, we can find the value of lambda as minus b plus minus b square. Minus b plus minus under root of b square. b square is this one 64 minus 4ac, that is 32 upon 2a2. Yeah. So this will be minus 8 plus minus 64 minus 32, that will be 32. 32 under root can be written as 64 into 2 means 4 root 2 upon 2. Okay. So our lambda is coming out to be minus 4 and plus minus 2 root 2, right? This is our lambda. This is our lambda. Now once we find the lambda, okay, once we find the lambda, once we find the lambda, what will be the center of this circle basically? The center of this circle will be minus g minus f, that is minus lambda by 2 and minus lambda by 2, right? So if you put the value of lambda here, this center will be basically this limiting point, right? So what I am doing, I am putting the value of lambda here. So minus lambda by 2 will be basically equal to this thing 4 plus minus 2 root 2 upon 2, right? That is equal to 2 plus minus root 2, right? This will be the limiting point. This will be the limiting point. But nowhere in option, it's mentioned, okay? Minus root 2 plus root 2 is there, minus root 2 minus root 2 is there. But our limiting point will be basically this, 2 plus minus root 2, okay? This will be our limiting point. This will be our limiting point. So option B, none of this is correct, right? So I think my laptop's battery is down. Let me check. Let me check this. Just give me a minute, okay? Sorry for the interruption. Actually, laptop's battery is drained almost. Okay. So we have done this question now. Question number 11 is done, I think. Yeah. Now let's take this next question, question number 12. So it's saying the limit, the point this is a limiting point of a coaxial system of circles of which X square plus Y square is equal to 9 is a member. The coordinates of the other limiting point is to be determined, right? So coaxial system of circles, one of the member of this system is given as this one, X square plus Y square, right? X square plus Y square minus 9 equal to 0. And other member, we can take this point, this point circle. This limiting point is given there, no? So limiting point, if you say limiting point is given as 2 comma 3. So we can, you can take this as a point circle having radius 0. So our second circle to this family, this family of coaxial system of circles, I can write it as X minus 2 whole square, right? X minus 2 whole square plus Y minus 3 whole square is equal to 0. Okay. So this one is our first member and this one, this S2 is our second member. And now I have to find the coordinate of the other limiting point, okay? So what will be the family of circles of this coaxial system? It will be basically S1 plus lambda times S2 is equals to 0, right? So the equation of S1, sorry, this will be X square plus Y square minus 9 plus lambda times S2. What is S2? X square plus Y square, right? And from here we get minus of 4X, right? Then minus of 6Y and plus 4 plus 9, that is 13 I think, right? So plus 4 and 9, okay? So plus 13 is equals to 0. Now what I can do, let me simplify it, okay? So I will get this thing, 1 plus lambda X square, right? 1 plus lambda X square, then 1 plus lambda Y square, 1 plus lambda Y square, then minus 4 lambda X, right? Minus 4 lambda X minus 6 lambda Y, minus 4 lambda X minus 6 lambda Y, then plus of 13 lambda minus 9 is equals to 0, okay? So I have to make, I have to divide this whole equation by 1 plus lambda to make the coefficient of this X square and Y square as 1. So dividing it by 1 plus lambda throughout, I will get X square plus Y square, right? Then plus of, what is there? Minus 4 lambda upon 1 plus lambda, okay? X then this minus 6 lambda upon 1 plus lambda into Y plus 13 lambda minus 9 upon 1 plus lambda, okay? This is the equation coming out, okay? Now if you see, if you see here, what is the value of G basically? G is coming out to be minus of 2 lambda, right? Upon 1 plus lambda and what is F? F is minus of 3 lambda upon 1 plus lambda and say this is as it is. So what I will do now, what I will do now, I will basically, yeah. So what I am going to do, I will, for finding the limiting point, what I will do? I will make the radius of this circle as 0, okay? So this thing G square plus F square minus A will be equal to 0. Under root of this will be 0, I am squaring both sides. So for limiting point, I am writing here. For limiting point, radius must be equal to 0, okay? So what is G square basically? G square will be this 4 lambda square, right? 4 lambda square, this 2G, I have done something wrong. This 2G is equals to minus 4, okay? Then okay, it's correct only, no? So 2G is minus 4 lambda, so G will be dividing by 2, that will be minus 2. Okay, okay, we are correct only. So G square will be 4 lambda square upon 1 plus lambda square, okay? And what will be F square? It will be 9 lambda square, 9 lambda square upon 1 plus lambda square. And what is C? Minus of C, minus of C means minus of this thing. So minus of C, what is C? This 13 lambda minus 9 upon 1 plus lambda and this must be equal to 0. So take this 1 plus lambda pole square as you will see. We get this 4 lambda square, okay? Plus 9 lambda square and this minus 13 lambda minus 9, this will be multiplied by 1 plus lambda, okay? So it will be 13 lambda square and what will be there here? This 13 lambda plus 13 lambda plus 13 lambda square, right? 13 lambda square, then we get minus 9 and minus 9 lambda, is it okay? I think it's correct. So this 13 lambda square minus 13 lambda square will be cancelled and we are left with finally minus of 13 lambda, okay? Minus of 13 lambda plus of 9 and plus of 9 lambda is equal to 0. So minus 13 plus 9, that will be 4 lambda. Minus of 4 lambda is equal to minus of 9. From here we get lambda is equal to 9 by 4, okay? From here we get lambda is equal to 9 by 4. And what is required? The coordinates of other limiting points, so center. So limiting point is nothing but the center of the point circle. So what will be center? Center will be minus g that is 2 lambda upon 1 plus lambda and this 3 lambda minus means 3 lambda upon 1 plus lambda, is it okay? This will be the limiting point of the center. Since limiting point will be center of the circle. So put the value of lambda here. So it will be 2 into lambda that is 9 by 2, right? 9 by 2 upon 1 plus lambda. 1 plus lambda will be 4, 13 by 4, right? 13 by 4. What will be this? 3 times lambda will be 27 by 4. 27 by 4 and 1 plus lambda 13 by 4. So finally we will get this 9 by 2 into 2 that is 2, 18 by 13, right? Yeah, 18 by 13 and it will be 4, 4 will be cancelled out, 27 by 13. So I am saying this option A, option A is matching with our answer. So this will be our answer. So the other limiting point will be this, option A, okay? So this was our question number 12. Let's check this next question. It is subjective, no options are provided here. So okay, two circles are drawn through the points this and this to touch the y-axis, okay? Prove that they intersect at angle this, okay? So two circles are drawn through the point this. We can treat it as the intersection point of both these circles, okay? And the circles are touching y-axis, okay? Prove that they intersect at, like we have to find the angle between both the circles, intersection point, okay? So what is basically angle? Angle of intersection means angle between the tangents at the point of intersection, okay? So let me assume, let me assume the equation of circle as this thing. X minus, let the equation of circle be X minus H whole square, right? X minus H whole square plus Y minus K whole square is equal to R square. Now since the circles are intersecting at y-axis, no? So the X coordinate, X coordinate of the center will be equal to radius X coordinate of center. That means this X minus H whole square, okay? Plus Y minus K whole square should be equal to this thing, H square, why? Why? Because it is touching y-axis. Hence this thing will be replaced like the X coordinate. The X coordinate of the center that is H here will be equal to the radius of the circle. So we can write in this way. Hope this is clear to everyone. Now let's expand this. So this will be X square plus Y square, right? Then minus 2 HX minus 2 KY. And what we get? Here we are having H square plus K square, right? Is equal to H square. Now this H square and H square will be cancelled out. So finally we got the equation of circle as X square plus Y square minus 2 HX minus 2 KY. And plus K square is equal to zero. Okay? So this is what we got the equation of the circle. This is one. Now it is saying that the circles are passing through this points A comma 5A and 4A comma A, right? So it must satisfy, it must satisfy, passes through what is given A comma 5A and 4A comma A, right? 4A comma A. So let's put here these values, these coordinates in this equation. We get this A square, okay? Plus Y square means 25A square, 25A square minus X is equal to A0. So minus 2 AH, then minus 2 Y is 5. So 10 AY. It will be basically 10. So it will be minus 2 into 5, 10, 10 AK, right? Minus 10 AK. And plus K square is equal to zero. So this will be basically 25 and plus 1. That is 26A square, right? 26A square minus 10 AK plus K square is coming out to be 2 AH, right? This is when I have put this point in the equation. Now let's take the second point in the equation and write the equation. We will get 16A square. X square means 16A square plus A square, right? Plus A square. X, place of X, I have to put H for it. So 28A, 8AH, right? Minus 8AH, then minus 2 AY, minus 2 AK, sorry, minus 2 AK plus K square is equal to zero. Now simplifying this, we will get 17A square. Minus 2 AK, 17A square minus 2 AK plus K square, okay? Is equal to 8AH. This is our second equation. So if we substitute the value of this AH node in any one of the equation, I think we will be having this quadratic in K. And from there, we have to find the value of K. Okay, so from 1 and 2 if you see, from 1 and 2, from 1 and 2 if you see, this will be 26A square, 26A square minus 10 AK, okay? Plus K square, this is the value of 2AH will be equal to this whole divided by 4. Minus 17A square minus 2 AK plus K square, this divided by 4, right? So basically it will create a quadratic in K, right? We will get quadratic in K, right? We will get quadratic in K and we have to find the value of K. That means it will be like linear equations. So let me check, I have already done this. So not to spend much time here. Let me check what is the value of K coming out to here. So like I am not going to solve, okay? So please cooperate on this. This is like quadratic in K. Okay, quadratic in K and I have got the value of K as, let me write from the copy. So K is coming out to be 1 value is 3A, okay? And the another value is this 29A upon 3. 29A upon 3, okay? Now what I will do, I will substitute the value of K here in equation 1 and 2. Like put value of K, put the value of K, okay? In equation 1, in equation 1 and 2, okay? So we will get the value of H, right? We will get the value of H. So our value of H will be 200. H will be 5A upon K. One value of H will be 5A upon 2 and say the another value will be 205, I think. 205A by 18, 205A by 18, okay? So this is what we got and we know the value of this, what do you say? K is known to us, no? So H and K, these things are known to us. Now what is H and K? H and K is the center of this circle. So basically H and K is the center of this circle. So like let me draw. So one circle is here, okay? We are having one another circle. So this one is the center of this, suppose C1, this is our C2, right? So the coordinates of C1 will be H, K. So this will be 5A upon 2 and 5A upon 2 and 3A, right? And what will be the coordinates of this C2? It will be H, K means 205, 205A upon 18 and comma 29A by 29A by 3A, okay? Now we have to find the point of, okay? They intersect at this angle, means we have to find this thing. Angle between the tangents drawn at the point of intersection. So if you see, this will be one tangent, this will be a separate tangent, right? And this will pass through this center only, right? So suppose I am taking this point, I am, suppose I am taking this point as the point of intersection, this 4A comma A, right? So suppose this one is our slope M1 and this slope is M2. This will basically pass through the center only, right? This lines will, these tangents, this radius will pass through the centers only. So C1 and C2. So M1 will be basically Y1 minus Y2. That means this 3A, 3A minus A, or you can write A minus 3A. Anything you can write, A minus 3A, Y1 minus Y2 upon X1 minus X2, that is 4A minus, 4A minus, what was the value? 5A upon 2. And what will be M2? M2 will be A minus 29A upon 3. And X2, this 4A minus 205A upon 205A upon 18, right? So the angle between that will be, that 10 theta will be coming out to be M1 minus M2 upon 1 plus M182. So it is not possible for me to like solve this much, okay? It means I have to devote much time on this single problem. So I'm leaving this to you guys, okay? Please solve it. I think we will be able to find the answers. Like I have worked it out in graph earlier, okay? So we will get the answer. So our theta will be coming, what is required in the question. So let's try from your end, okay? So I'm leaving this up to you. Still we are having, I think, four, five problems in this exercise. So let's focus on that, okay? So it is saying this question number, 14. Find the equation of a circle which cuts orthogonally the circle this and passes through and touches the axis of Y. So one equation, one circle is given here, right? One equation, one circle is given here. Let me write it as S1. Let me change the color. So S1 is given as X square plus Y square minus 6X plus 4Y, right? 4 plus 4Y minus 3 equal to 0. And let me assume this is another circle as let S2 be, let S2 be X square plus Y square plus 2GX plus 2FY plus equal to 0, okay? Now this circle is passing through, passes through, passes through 3, 0. Then it should must satisfy. So this will be basically 9, right? 9 plus 0 and 2G that is 6G plus 6G. Again, this will be 0 plus equal to 0. So basically we got 6G plus 9 plus C equal to 0, okay? This one equation we got from passing this circle through this point, okay? Now one more thing is given here. Like these two circles, this S1 and S2 are orthogonal, right? This S1 and S2 are orthogonal. So from here also we will get something. So if you see S1, if you see here, for S1 what is the value of G1? So what will be the value of G1 basically? It will be minus 3, right? And what is the value of F1? Value of F1 will be 2, am I right? Okay. So this thing will be 2G1G2. Means condition, I am writing the condition of orthogonality. 2G1G2 plus 2F1F2 is equal to C1 plus C2. So 2 times G1 is minus 3. What is G2? G2 is nothing but G1. And 2F1 is 2. And what is F2? F2 is F. What is C1? C1 is minus 3, right? And our C2 is plus 6. So here we get minus times minus of 6G, right? 2, 3, 6. Yeah, minus of 6G plus 4F plus 4F is equal to C minus 3. Okay. So I think we can add both these equations. Okay. So it will be basically minus 6G plus 4F minus C plus 3 equal to 0. Okay. Now I will add 1 and 2. Like 1 plus 2. So what we will get? This 6G thing will be cancelled out. So 9 plus C, right? 9 plus C plus 4F minus C plus 3 equal to 0. I have omitted this G term. 6G minus G will be cancelled. Here the C thing is also getting cancelled out. The C is also getting cancelled out. So we got 4F plus 4F plus 12 equal to 0. So we got F as minus 3. Okay. We got F as minus 3, right? So if we got F as minus 3, what can we do with that? F is minus 3G. Okay. One more information is with us that the circle, find the equation of circle, this circle, this S2 is touching the axis of Y, right? So it's radius will be basically, it's radius will be, like this will be the condition now. Let me draw for better understanding. So like this is our Y axis. So this is our Y axis. Sorry. This is our Y axis, no? And it's center is what? Minus G comma minus F. This is our S2 circle. This center is minus G minus F. So basically if you see this under root of G is square plus F is square minus A. This is the radius and this radius is nothing but if it is touching the axis of Y, it will be equal to minus G only. Or we can say mod of G. So square it. After the squaring, we will get G is square plus F is square, F is square means 9. And minus of C will be equal to this G is square, right? So G is square and G is square will be cancelled. We got the value of C as 9. So we got the value of C as 9. Now once we got the value of C, we can put in equation 1. We can put in equation 1 and we get it as 6G plus 9 plus 18. That is a plus 18 will be equal to 0. So from here we got G as minus 3. Am I right? Minus 18 by 16, yeah. So finally the equation of circle will be, finally the equation of circle will be this thing. X is square, okay? Plus Y is square plus 2 into G. This is 2GX plus 2FY and plus C. C is nothing but 9 equal to 0. That means X is square plus Y is square minus 6X minus 6Y. And plus 9 equal to 0. Is it okay? I think this will be the answer. Minus X, minus 6Y plus 9 equal to 0. Yeah. This will be our answer. So we are done with this question. Let's move to question number 15. So tangents are drawn to circle this. From any point on the line this, prove that their lengths are equal. Prove that their lengths are equal, okay? So like this is like this condition. One circle is given here, okay? One another circle is here. And one line is there, okay? One line is there. And it is saying the length of the tangent is drawn, okay? The length of the tangent suppose this thing. So I'm drawing one line, one tangent from here. And I'm drawing one tangent from here, okay? So this is what it is given. This line is basically 2X, okay? 2X plus 3Y minus 5 equal to 0. This is our first circle. Suppose this is our X square plus Y square plus 4X plus 6Y minus 19 equal to 0. And this is our second circle. X square plus Y square is equal to 9, right? So it is saying that this PA, suppose this point is B, this point of tangency is A, this point of tangency is B, okay? So if PA, like, prove that their lengths are equal, okay? PA and is equal to PB. This is what it is saying. So if PA is equal to PB, what does it mean? The length of tangent from this line, from any point on this line to these two circles are equal. That means this 2X, this 2X line, line 2X plus 3Y, right? 2X plus 3Y minus 5 should be radical axis, right? Should be radical axis of the given circles. Should be radical axis of the given circles, of given two circles. So our target is to prove anyhow that this line is a radical axis of these two. So what will be the radical axis? If you see, radical axis of this, suppose this one is S1, this one is S2, radical axis of S1 and S2. So it will be S1 minus S2 equal to 0. That will be nothing but 4X plus 6Y minus 19 minus time, minus of minus 9 equal to 0. That will be 4X plus 6Y minus 19 plus 19, that will be minus 10 equal to 0. So 2X plus 3Y minus 5 equal to 0, right? So this is the radical axis. This is radical axis, the line given, given line, given line. So length of tangents will be equal. Length of tangents will be equal. This is the definition only. This is the definition of the radical axis, right? So radical axis is the line from where tangents drawn to the given circle will be equal. The length of the tangents will be equal. So hence it is proved basically. So we are done with this question. Let's move to the next question, question number 16. So it's saying to find the coordinates of the point from which the length of tangents to the following three circles will be equal. Find the coordinates of the point from which the length of the, okay, like first we have to find, the question is asking for, question is asking for radical center. Okay. So we have already done this type of questions in this particular exercise only. So I will leaving this to you guys. Okay, you have to solve. Now how we can solve S1 is known to us, like three circles are given here. S1, S2, S3, right? So we can find the radical axis one, radical axis one of S1 and S2. That will be nothing but S1 minus S2 equal to zero, right? And radical axis two of this S2 and S3 can be found out by S2 minus S3 equal to zero. Okay, now take this, take intersection point of these two radical axis. So we can find the, take intersection point of these two, these two radical axis. So we will get radical centers. Okay. So hope you guys can do it. Now let's move this, let me move to this next question. It is saying find the equation of a circle which is coaxial with this and this and having its center on the radical axis of these circles. So it is saying this, suppose this is our first circle. What? This x square plus y square plus 4x plus 2y plus 1 equal to zero. Okay. And the equation of second circle is basically x square plus y square minus x plus, plus 3y minus 3 by 2 equal to zero. Okay. Now we have to find the equation of circle which is coaxial with these two circles and having its center on the radical axis of these circles. Okay. So basically if you see what will be the circle which will, which is coaxial with these circles. So it will represent the family of circle with equation s1 plus lambda s2, right? s1 plus lambda s2. Am I right? So this will be the required equation. So let's try it, let's put the value x square plus y square. This plus 4x plus 2y. Okay. Plus 1 and plus lambda times s2. That is nothing but x square plus y square minus x plus 3y minus 3 by 2 equal to zero. Okay. So it will be basically 1 plus lambda x square. 1 plus lambda x square. Then 1 plus lambda y square. Okay. Then 4 minus lambda, right? 4 minus lambda x. Then 2 and here 3, right? 2 plus 3 lambda y. 3 lambda will be from here. 2 will be from here. So 2 plus 3 lambda y and plus 1 minus 3 lambda by 2, right? So make the coefficient of x square and y square as 1. We will have x square plus y square plus 4 minus lambda upon 1 plus lambda into x plus 2 plus 3 lambda upon 1 plus lambda, right? Into y and 1 minus 3 lambda by 2 upon 1 plus lambda is equals to zero. Okay. This will be the thing coming out. Now what is the center of this? If you say G will be, what will be our G? G will be 4 minus lambda upon 2 into 1 plus lambda for this. And what will be F? F will be 2 plus 3 lambda upon 2 times of 1 plus lambda. Okay. And what will be center? What will be center? Center will be minus G. That will be minus means lambda minus 4 upon 2 into 1 plus lambda. We can write this way. And F will be minus of F, no? So minus of 2 plus 3 lambda upon 2 times 1 plus lambda. Now it will lie on the radical axis of these two circles. So what will be the radical axis of these two? Radical axis will be S1 minus S2 equal to zero, right? So put it. So this will be basically 4x minus of x. That will be 5x. And 2y minus of 3y will be minus of y. And 1 minus of 3, that will be plus 5 by 2. Plus 5 by 2 equal to zero. Okay. Now this center should satisfy. This center should satisfy. Center should satisfy this equation, right? Center should satisfy this equation. So after putting the value, we have the value of lambda, right? We get the value of lambda. Now put that value of lambda. Put the value of lambda. Okay. Put the value of lambda in this equation. In this equation 1. Put the value of lambda in equation 1. We will get the value of, means we will get the equation of circle. Is it clear? The question is saying that the, what do you say? The center of this circle, the center of this circle is lying on the radical axis of this circle. So I have identified the radical axis here. We have got the radical axis of the required circle. Sorry. We have got the center of the required circle. So this center should satisfy this equation. This equation of radical axis. So putting the value of x and y in terms of lambda here, we will get the value of lambda. Now put that value of lambda in this equation. We will get the required answer. Oh, this is clear to all, right? So I think one more. This is the last question. So let's finish it off. Find the radical axis of a coaxial system of circles whose limiting points are this. Okay. So limiting point is given as, this limiting point 1 is given as 1, 2. And our limiting point 2 is given as 3, 4. So what are these? These are basically point circles, right? These are basically point circles whose radius is, whose radius is equal to 0. So let me write the equation of circle, point circle. x minus 1 whole square, right? x minus 1 whole square plus y minus 2 whole square is equal to 0. And what will be the second circle? It will be x minus 3 whole square, then plus y minus 4 whole square is equal to 0. Now what we need to find? We need to find the radical axis. Okay. So this will be basically x square plus y square. From here we get minus 2x, then minus of 4y plus 1 plus 4, that is 5, plus 5 equal to 0. And what will be our S2? S2 will be x square plus y square minus 6x from here, minus 8y from here, plus 9 and plus 16, that will be 25, right? So our radical axis will be, radical axis will be S1 minus S2 equal to 0. So this is minus 2x minus 4y plus 5 minus of, this thing, no, minus 6x minus 8y plus 25 equal to 0. So 6x minus 2x will be 4x, then 8y minus of 4y will be 4y. Here from, you get minus 25 plus 5. So this is minus 25 and plus 5, that will be minus 20, minus 20 equal to 0. So take 4 common, like divide it throughout by 4, we will get x plus y minus 5 equal to 6. So this will be the radical axis of circles whose limiting points are this. So guys, we are finished with this chapter, right? I think this was the last exercise of circles. So circle is also completed. We have done with all the exercises of this circle. So soon we will meet with next chapter, most probably next chapter of our conic section only. So let's see. Till then, Tata, goodbye. Take care, okay? So bye-bye.