 So, good morning, students. So we were doing the exercise five of circles in which I think I have completed up to a question number eight, right? So, we will start with question number nine today and we will finish this exercise today. Right. So, let's see question number nine. It is saying that the length of tangent from. It means zero comma zero to the circle. This is okay. So, equation of circle is given here as two X square. Two into X square plus Y square plus X minus Y plus five equal to zero right and it is asking for the length of tangent from zero comma zero. So, we know the length of tangent, the length of tangent is given by under root S one where S one is the power of the given point, right? But for this formula, we need to make the coefficient of X square and Y square as one right then only this formula is applicable. So, what I will do, I will make the coefficient of X square and Y square equal to one for that I have to divide this whole equation by two. So, dividing that, what we get X square plus Y square plus one by two X minus one by two Y plus five by two is equal to zero, right? Now, we can apply formula. Now, we can apply this formula, what we know that is the length of tangent is equal to under root of S. So, put the value of this, what you say this coordinate here. So, putting zero comma zero here, zero comma zero in this equation of the circle. We get this will be zero zero zero zero plus five by two. This will be the length of tangent, right? Length of tangent from zero comma zero. That two, that will be under root, right? So, it will be coming up to be under root of five by two, okay? So, this option B is correct. Option B is correct. Now, let's take the next question, question number 10. It is saying the perpendicular tangents to the circle X square plus Y square is equal to A square meet at P, okay? Then the locus of P has the equation. Okay, means the angle between the tangents are 90 degree, right? Angle between the tangents are 90 degree. So, actually this will, the locus of P will resemble the director circle, right? Because the tangents are meeting at 90 degree, okay? Let me first draw the sketch for this. So, like, this is our point P, okay? This is our point P, where this and the angle between them are 90 degree. So, if you remember, when the tangents meet at 90 degree, so this locus of P, suppose I'm taking its coordinate as h comma k, the locus of P is basically the director circle. It's basically this locus of P will be director circle. And what is the director circle? Means what is the equation of the director circle? Basically, director circle is co-centric, like on, it has same center as the given circle. What is the center of this circle, given circle? This is 0 comma z. So, the center of director circle will be same as the center of the given circle. But the radius of director circle is basically root 2 times the radius of the given circle. So, we can simply write the equation of director circle as x square plus y square is equal to, what is radius of given circle here? A, so it will be root 2A. This will be the radius of the director circle. So, from here, we get x square plus y square is equal to 2A square. This will be the locus of P. So, if you have an idea of the director circle, you can simply write the locus of P. But let's see like how this thing arrives, like how this locus of P arrives. So, it's simple basically. If you see, I'm taking the coordinates of P as h comma k, right? So, what I will do, this is our center of the circle, right? This is our, this is center of circle, which is nothing but 0 comma 0, okay? What I will do? I will join this P to C. I will join these two points, P and C, okay? And I will drop perpendicular from C on these tangents also, okay? So, if you notice this line P, C, this line joining P and C, you actually bisapes this angle. It actually bisapes this angle means what? This angle will be 45 degree, right? And this angle will also be 45 degree. So, if you see, I'm naming it as A and B. So, if you see in triangle P, A, C, if you see in triangle P, A, C, okay? 1045 degree will be equal to AC upon, AC upon, this will be 90 degree nu. So, 1045 will be AC upon PA, right? And what is AC? AC is nothing but the radius of the circle, which is A here in this case. So, AC, we can write AC as A and what will be PA? PA is the length of tangent, right? This is the length of tangent from P. So, this will be under root of S1, okay? So, we can say this under root of S1 is equals to 1045 is 1. So, under root of S1 is equal to A. And what will be under root of this S1? This will be nothing but h square plus k square minus A square, right? This whole thing will be under root and this will be equal to k. So, squaring both sides, what we get? We get h square plus k square minus A square is equal to A square, okay? Now, replace this h and k by x and y. So, finally, we get this x square plus y square is equal to 2 square, right? This will be the locus of given point P, which I told you earlier only. This is the equation of director circle. This is what we used to call this locus of P. We used to call this as director circle because y, because the equation is coming. As the circle only and the radius of that circle will be root 2 times the radius of the given set. So, this will be our answer. So, in option if you see, this option A is correct. Option A is correct, okay? Now, let's move to the next question. Question number 11, okay? It is saying the tangents to x square plus y square is equal to A square and inclinations alpha and beta intersect at P, okay? Equation of circle is given and it is given that the pair of tangents intersect at P. If cot alpha plus cot beta is equal to 0, then the locus of P is, okay? So, same thing we will draw once again. So, this is our circle. This is our circle. Okay, this is our point P. And we have to find the locus of point P, right? So, this is our point P. I am taking its coordinates as h from a k. This is our circle x square plus y square is equal to A square. And this inclinations, suppose the inclination of this line is alpha. This is basically with respect to positive x axis, right? So, we can say that it's a slope. The slope of this line is basically tan alpha, right? Since inclination is given as alpha. And its inclination, the line of the inclination of this line is given as beta. So, it's the slope will be basically what tan beta, okay? So, if you see, if I have to write the equation of tangent from P, okay? On this circle, how can I write it? I can write it as y is equal to mx plus minus c. C is nothing but A into 1 plus m square, right? This is the condition for tangency, no? This is the condition for tangency where c square must be equal to A square into 1 plus m square. This is the condition for tangency, okay? So, I am writing the equation of tangent as y is equal to mx plus minus A into 1 plus m square, right? Because the two values of c will come from here, plus minus A under root 1 plus m square. Now, this line is passing through h comma k, right? So, I can replace this x and y by h comma k. So, we can write the equation, rewrite the equation as k is equal to mh, right? Plus minus under root A, sorry, plus minus A under root 1 plus m square. This is the equation of tangent, right? Equation of tangent, which is passing through P, okay? Now, let's make it a quadratic in m. So, what I get? I will be having k minus mh whole square is equal to A square into 1 plus m square, right? So, open it, it will be k square plus m square h square minus 2 khm is equal to A square plus A square m square, okay? So, now, let's take the coefficient of m square together. We get this h square minus A square into m square, okay? Then, I will get minus 2 khm into m, okay? And then, plus k is square minus A square, okay? So, we got quadratic in m. What we got? We got quadratic in m. So, obviously, from here, we will get the two values of m, means we will get two values of m, right? From here, if you see, our m1 plus m2, our sum of roots for this quadratic will be minus B upon m. That is nothing but 2 k h upon 2 k h upon h square minus A square, okay? And what is m1 and m2? We are given the slope of tangents, right? Here, the slope of tangents, if you observe, we are provided with the slope of tangents in the question itself. So, we can replace this, we can write this m1 plus m2 as 10 alpha plus 10 beta, right? So, the value of 10 alpha plus 10 beta is coming out to be this 2 k h upon h square minus A square. Now, one more information is given here, this cot alpha plus part beta, okay? So, let's make it in terms of cot alpha and cot beta. I will get 1 by cot alpha plus 1 by cot beta is equals to same this thing, 2 k h upon h square minus A square. So, taking LCM, what we get? cot alpha plus cot beta upon cot alpha into cot beta is equals to 2 k h upon h square minus A square. Now, this thing is given as 0, right? This is given as 0 in the question itself. So, from here, if you see what we get, we get 2 k h is equals to 0, right? Now, for locus, we know that we replace this h and k by x and y. So, replacing this, what we get? 2 x y is equals to 0 or we can say x y is equals to 0. So, this will be the locus of point p. This will be the locus of point p. So, x y equal to 0. This way option C will be correct. Hope this is clear to all. This is a good question, basically. So, let's take the next question, question number 12. It is asking the exhaustive range of values of A such that the angle between the pair of tangents drawn from A comma A to the circle lies in the range pi by its way to pi. Okay, means one circle is given here. And we have to draw a pair of tangents, but the angle between the pair of tangents is limited to pi by 3 to pi. Then we have to find the value of A. Obviously, it will lie within some range. So, let's see how can we do it. So, I'm drawing this purple. Okay, and from point A, we have to draw a pair of tangents, right? From point A, we have to draw a pair of tangents. This angle is given. So, I'm doing one thing. I'm joining the center also because later on we have to do this construction and I'm joining these points. So, this is all. Let me name it as point P. Okay, whose coordinates are given as A comma A. Let me name this point of contact as AB, this center as C. Okay. So, given, if you see the equation circle, what will be the coordinates of this C? It will be 1 comma 1, right? It will be 1 comma 1. And what will be the radius of the circle? It will be 1 square plus 1 square minus C minus A is means minus of minus X that will be plus 6. So, radius will be 1 plus 1 to under root 8. The radius of this circle is under root 8 and center is 1 comma 1. Now, it is saying that the angle between the pair of tangents should lie in the range pi by 3 to pi. Okay. So, suppose I'm taking this angle as 2 alpha, right? So, this 2 alpha as per given condition, this 2 alpha should lie between pi to pi by 3. Okay. If this angle, this angle between pair of tangents is 2 alpha, what will be this angle? This angle will be alpha, right? Because this PC will bisect this angle. So, this will be alpha, this will be alpha. Now, we will play with this alpha means, so, if 2 alpha is lying between pi by 3 to pi, where should this alpha lie? Alpha should lie between pi by 6 to pi by 2, right? And if I say, where should sine alpha lie? Sine alpha should lie between, what is sine pi by 2? It is 1 and what is sine pi by 6 means what is sine 6, sorry, sine 30 is 1 by 2. Okay. So, this will be the range of sine alpha as per given condition. Sine alpha should lie between 1 by 2 to 1. Now, what is sine alpha? From this figure if you see, what is sine alpha? Sine alpha will be, this will be a right angle triangle, right? So, sine alpha will be basically AC, right? AC upon PC. Okay. And what is AC? AC is nothing but the radius of the circle. So, I am representing at R and what is PC? We can find this length of PC since coordinates of A and C are known to us. So, if you see, I am writing here, PC will be nothing but under root of A minus 1 whole square, right? Plus A minus 1 whole square. So, this is nothing but 2 times, 2 times A minus 1 whole square under root. Okay. So, this will be under root of 2A minus 1 whole square. Okay. Now, sine alpha is lying between 1 by 2 to 1, right? So, I can say this thing, put the value of sine alpha here. So, let me write in terms of AC and PC only because it will be short. So, R upon PC should lie between 1 by 2 to 1, right? From here, if you see, this is an inequality where we can see, multiply it by 2. It will be 1, 2 R upon PC should be 2. So, from here, we get two conditions like this PC should be greater than R and PC should be greater than R and 2R is greater than PC, right? Or PC is less than 2R. So, PC is greater than R, but it should be less than 2R. Okay. Now, what is PC? PC, we know the value of PC, length of PC. So, it will be under root of 2A minus 1 square should be greater than R. Okay. So, square it, we will have two times A, A square minus 2A plus 1 should be greater than R. R squared will be 8. Okay. So, this will be cancelled out. So, finally, we get A square minus 2A plus 1 minus 4 means minus 3 should be greater than 0. So, factorize it. It will be A square minus 3A plus A minus 3 greater than 0. So, A, A minus 3 plus 1, A minus 3 should be greater than 0. Finally, we get it as A plus 1 into A minus 3 should be greater than 0. Okay. Now, find the solution for this inequality. So, let me draw the critical points. It will be minus 1 and plus 3, right? So, anything value of A greater than 3 will give me positive result. This will be negative and this will be negative, but what we need, we need positive. So, we get the value of A as minus infinity to minus 1 and union 3 to infinity. But our work is not yet done. We have to solve this inequality also. PC is less than 2R. So, solving this, if you see, to under root of A minus 1 whole square, right? This is PC. This should be less than 2 times R, that is 2 under root 8. So, cancel it. Oh, no. This will be under root, no? Oh, I have written it wrongly. So, this will be, let me erase it first. So, what I have to do? This PC is nothing but under root of 2A minus 1 whole square, right? This is PC. This should be less than 2 times under root 8. Now, square it. So, we get 2 into A square plus 1 minus 2A should be less than 4 into 8, that is 32, okay? So, it will go 16 times. So, finally, we get quadratic as A square minus 2A plus 1 minus 16, that is minus 15. This is less than 0. So, split it, A square minus 5A plus 3 minus 15 should be less than 0. So, A into A minus 5 plus 3A minus 5 should be less than 0. So, finally, we get A minus 5 into A plus 3 should be less than 0, okay? Plot the, now plot that critical point here for this equation, okay? What we get? This will be minus 3. This will be our plus 5. So, anything ever 5 will give me positive result. So, this will be positive. Anything between minus 3 to minus 5 will give me negative and anything less than minus 3 will give me positive result. But we need negative. So, from here, we get A belongs to minus 3 to 5, right? Now, we have to take the intersection of both these solutions, right? We have to take the intersection. So, finally, we get AS, finally, we get AS, A belongs to from minus infinity to minus 1. So, minus 3 to minus 1, right? And union, here it is going from 3 to infinity, but limiting point is 5 only. So, it will be up to 5 from 3 to 5. So, this will be our final range of A for which conditions such that the angle between the pair of tangents will be in the range of pi by 3 to pi. So, this will be our answer. Option, option, if we take the option, option B is the left, okay? So, this was our question number 12. Now, let's move to the next one. Okay. So, it is seeing distances from the origin to the centers of the three circles, x square plus y square minus 2 lambda x equal to c square, where c is constant and lambda is available. No, it should, it should be variable basically. So, let me correct it. So, it is saying that lambda is variable, variable. Okay. So, a lambda is variable, R in GP means distances from origin to the centers of circle are in GP, okay? So, I think that the lengths of tangents drawn from any point on the circle, x square plus y square is equal to c square to the three circles are also in GP. Okay. So, I think question is clear to everyone. Like distances from origin to the centers of the circles. Let me write the equation of the circles. So, write the equation of first circle. So, it will be like x square plus y square minus 2 lambda 1x since lambda is variable. So, for circle 1, I'm taking lambda is lambda minus x minus c square is equal to 0. Similarly, we can write the equation of circle 2 and x square plus y square minus 2 lambda 2x, okay? So, minus c square is equal to 0. Rest all things with c will be constant since it is constant. And we can write the equation of circle 3x, x square plus y square minus 2 times lambda 3x minus c square is equal to 0. Okay. So, tell me what will be the center for this circle. If you see what will be the center for this circle, it will be basically lambda 1, 0. What will be the center for this circle? Okay. What will be center for this circle? It will be lambda 2, 0. And circle for this center for this third circle will be lambda 3, right? Now, distances of these points, distances of these points from center is constant, not constant, RNG. Okay. Similarly, let me assume this origin as O. So, this O c1. O c1 will be lambda 1 minus 3 to the square means distance formula. So, it will be under root of lambda 1 square that is nothing but lambda. Similarly, our distance of second center means the center of second circle to origin will be under root of lambda 2 square that is nothing but lambda 2. So, distance of this center of third circle to origin will be lambda 3 square that is lambda 3. So, these are the distances as per question it is given that this lambda 1, lambda 2 and lambda 3 is in GP, right? Now, what we need to prove? We need to prove that the lengths of tangents drawn from any point on the circle. This given circle x square plus y square is equal to c square to the three circles are also in G. Okay. So, one circle is given here, x square plus y square, x square plus y square is equal to c square. Let me assume one point on this circle as, let me assume one point on this circle as p whose coordinates we can give it as c cos theta. r cos theta comma r sin theta, we used to take any point on the circle in parametric form in this way only. So, I am taking one point on this circle as c cos theta comma c sin theta, right? Now, I will take, means we have to draw the tangents from this point on the three circles, okay? We have to draw the tangents from this point to the three circles, okay? So, suppose I am taking it as, now I am calculating the length of tangents, okay? So, this length of tangents, I will calculate length of tangents from this point on the circles, okay? So, how can we do it? For first circles, let me denote it as under root of S1, okay? So, what will be our under root of S1? This x and y wherever this x and y are there in the equation of these circles, we will replace it that x and y by this c cos theta and c sin theta. So, it will become c cos theta whole square, right? x square plus y square, y we will replace it as c sin theta square, right? Minus 2 lambda 1 into c, sorry x, x is nothing but c cos theta, right? And minus of c square, this whole thing will be under root. So, if you see here, this will become c square cos square theta plus c square sin square theta. That will be nothing but c square, then minus 2 lambda 1 c cos theta, okay, minus c square. So, this c square, this minus c square plus c square will be cancelled out. And look at this, here I can assume this thing, this minus 2 c cos theta, okay? I am writing it separately and multiplied by lambda 1, okay? Now, observe that this will be some constant, right? What I have written in this bracket, it will be some constant, some constant k. Why? Because c is also constant, 2 is constant and this cos theta will also be constant because what I have done, I have taken one point on this circle, okay? Whose inclination with the x axis is theta? That is constant for this point. So, we can say it as some constant as k. So, this under root a1 is coming out to be under root of, under root of some constant, we can place it outside also. So, it will be nothing but this k into under root of lambda 1. So, this will be our length of tangent on the first, right? And similarly, the length of tangent on the second circle will be k into under root of lambda 2. And our length of tangent on the third circle will be k times under root of lambda 3, right? And we know that this lambda 1, lambda 2 and lambda 3 are in GP, right? It is given in the question. So, obviously, if lambda 1, lambda 2 and lambda 3 are in GP, right? So, this thing, this S1 means k into this lambda 1 or k into under root lambda 2 and k into under root of lambda 3. This will also be in GP, okay? You can take any values and you can also, this if lambda 1, lambda 2 and lambda 3 are in GP. Their roots will also be in the GP. So, we have proposed this. Hence, this is what we were asked in the question. Okay. So, now we will take this next question, question number 40. It is saying that they find the area of the quadrilateral formed by a pair of tensions from a point, from this given point, 4 comma high to this given circle and the pair of its radii. Okay. Same sketch we will draw once again. These two centers. And we will join this point. We will join this point to the center. Okay. So, let me name it as P, this point of contact says AB and this is the C center of the circle. Okay. Now the coordinates of this point P is given as 4 comma 5. And we know the equation of circle as X square plus Y square minus 4 times minus 2 Y minus 11 is equals to C. The center of the circle, it will be 2 comma 1. Right. So, we know the coordinates of this C as 2 comma 1. Okay. And what will be radius for this circle? It will be 2 square means 4 plus 1 minus C minus of minus 11. That is nothing but plus 11. So, it will be 4 plus 1 pi 11 plus 5. So, 11 plus 5, how much? Have I done some mistake or what? So, center of this circle will be 2 comma 1. 2 comma 1. Right. And it will be 12, 16. Okay. Radius is coming out to be 4. Okay. Route 16 is nothing but 4. So, radius is 4. So, we got this radius as 4. AC is out, nothing but radius only. So, the length of AC is 4. Okay. This will be a right angle triangle. So, question is asking to find the area of this quadrilateral PACB. Okay. But we already know that this line PC. Both this, if I write it as triangle PAC and triangle PBC, both will be congruent. Why? Because these angles are equal. These sides are equal. And this is the common side. So, side angle, side congruence, we can show here. So, anyhow, if we find the area of this triangle PAC, area of triangle PAC, and we will double it, then we will find the quadrilateral serial. So, area of triangle PAC, if you see, area of triangle PAC is equal to half in base, like this is a, what do you say PA, right? Half base into height, that is AC. Okay. And how can we find this PA? This is the length of tangent basically. So, from here, we can find PA. PA is under root of S1. We know the coordinates of PA. So, I will put there. So, this will be 4 is where that is 16 plus y is where that is 25, minus 4 into 4 minus 16, minus 2 into y, that is minus 10 and minus 11. So, how much it is coming? 16 and 16 will be cancelled out. This 25 minus 21, that will be 4, means 2. So, PA is coming out to be 2. Now, put the value of PA here. So, it will be half into AC. AC is our 4. So, this is 4. So, area of triangle PAC is coming out to be 4 square in it, right? So, our area of quadrilateral will be, area of quadrilateral, PACB. PACB will be 2 times 4, that is 8 square in it. Okay? So, we got the area of quadrilateral as 8 square in it. So, this was our question number 14. Now, let's take this question number 15. It is saying if the length of tangents from a given point, whose coordinates are f from a g to a circle this, before tense the length of the tangent from the same point to the circle this, x square plus y square is equal to 4x. Now, the question is asking to show that this, this, this is equal to 0. Okay? We know the point that this point P whose coordinates are f, we have to find the length of tangents from this point on these two circles. Okay? So, we know that length of tangent is under root of S1, that is under root of the power of the given point. So, it is given as under root of S1 is 4 times under root of S2, right? To the circle, we 4 times the length of the tangent on the another circle. So, what will be S1? If you see here, S1 will be basically, I will replace this x and y by f and g. So, S1 will be f square plus g square minus of 4. This will be our S1, okay? And what will be our S2? S2 will be f square plus g square minus 4 into minus 4 into x, x is f. So, this is what S1 and S2. Now, put the value there and square it. So, we get this S1 square is equal to 16 into 16 times S2, right? So, S1 square will be, sorry, this square will not come. This is already in the under root, no? So, S1 will be basically equal to 16 times S2, okay? So, put the value of S1 now. It will be f square plus g square minus 4 is equal to 16 f square. f square plus 16 g square minus 16 into 4, 64 f, right? Now, take all the terms to the one side. So, this 15, 16 f square minus f square, that will be 15 f square. 16 g square minus g square plus 16, 15 g square, right? And this minus 64 g plus 4 will be equal to 16. This was, this is what asked in the question. So, we have proved that, yeah, the value of 15 f square plus 15 g square minus 64 f plus 4 is coming out to be 0, okay? So, let's take this question. Find the equation of that part of the circle, which is bisected at 3 comma 2. So, one circle is there and its chord is given as, suppose one chord is given here, which is bisected at 3 comma 2, right? So, it's middle point. So, suppose this is AB. So, it's middle point, okay? This middle point, the coordinates of this is given as 3 comma 2. And now, we have to write the equation of this AB. We have to write the equation of AB, okay? Whose middle point is going to us, which is 3 comma 2. So, if the middle point of a chord is going, so we basically write the equation of AB as T is equal to S1. This is what the formula says, T is equal to S1. And what is the equation of circle? This is x square plus y square minus 15 is equals to 0, okay? So, what is T? Basically, normally, we write the equation of tangent putting the coordinates of this midpoint, okay? And what is S1? S1 is nothing but the coordinates of this point, it means power of this middle point. So, in this way, we find the equation of this AB. So, what is T basically? How do we write the equation of tangent? We replace x square by xx1, right? So, this will be xx1 and replace y square by yy1. So, yy1 minus 15 is equal to S1. Now, what is the power of point? We put the coordinates in the equation itself. So, x square means 3 square, y square means 2 square and minus 15, right? So, this is what we got. Now, x1 is equal to 3 for this and y1 is equal to 2. So, take it. So, this will be 3x, okay? 3x plus 2y minus 15 is equal to 9 plus 4 minus 15, okay? So, this will be 3x plus 2y minus 15. And what is this? 9 plus 4, 13 minus 15, that is minus 2. It will become plus 2 when it comes to this side. So, finally, we get the equation of abs, 3x plus 2y, sorry, this 3x, 3x plus 2y minus 13 is equals to 0. So, this is our equation of ab. This is what we were asked to find, okay? Now, let's take the, I think one more question is left in this exercise. Yeah, this question number 17. So, it is saying the quads of contact of the pair of tangents to the circle x square plus y square is equal to 1, drawn from any point on the line this, pass through the points alpha, comma, beta, then alpha square plus beta is broken. So, one similar question we have done in this exercise itself. This includes the concept of what you say, family of lines, right? So, this question is also based on the same concept. So, this is our pair of tangents, which we are drawing here, okay? And we have this point is lying on the given line, what is given in the question. This is lying on one line this, whose equation is known, it was, okay? So, let me take this point as p, okay? As usual, we are taking this point as p only and on the line, this equation of this line is 2x plus y is equal to 4, okay? So, this point p is lying on this line. Like this point is variable, this point p is variable, it can lie anywhere on this line, okay? So, after, like, how can we write this coordinates of p? If we say, we can rewrite it as this equation as y is equal to, right? y is equal to minus 2x plus 4, okay? You can write this equation, 2x plus y equal to 4 is given. So, we are writing this as y is equal to minus 2x plus 4. So, suppose if I am taking this point as h, the x coordinate as h. So, what will be the y coordinate? Y coordinate will be basically minus 2h plus 4. Is it clear? Because this point is p is lying on this line only. So, anywhere it lies on this line, the coordinates of this point p will be of this formula, h comma minus 2h plus 4. Okay? Now, question is saying this point of contact. Let me join this also. Okay? So, it is saying the codes of contact. Look here, it is saying the codes of contact. It will be not a single line, it will be a group of lines or a family of lines. So, the codes of contact, this is our code of contact, okay? Of the pair of engines to the circle. This is circle x square plus y square is equal to 1 drawn from any point on the line. Okay, we have taken that point, passes through point alpha comma beta. So, this code of contact should definitely pass some fixed point alpha comma beta. And we need to find that point. We need to find that point alpha comma beta. Then only we can find this alpha square plus beta square. So, let me write the equation of AB. How can we write the equation of AB? How can we write the equation of AB? P is equals to 0, right? Where the tangents, like the equation of tangents to be drawn assuming the point as this, coordinates of P. Okay? So, our equation is given as xx1 plus yy1 plus yy1 minus 1 equal to 0, okay? So, x into x1, what is our x1? That is h and y into y1, what is our y1? 4 minus 2h, right? 4 minus 2h minus 1 equal to 0. So, finally it becomes hx plus 4y minus 2h y minus 1 equal to 0. Okay? So, we can rewrite it as 4y minus 1. I am taking these two terms together. So, this will be 4y minus 1 plus h into x minus 2y, right? So, what does this represent? This is representing the family of these lines, means family of this chord of context, which are passing through the intersection of these two lines. Suppose I am taking this as l1 and this as l2. So, this family of lines will definitely pass through the intersection point of these two lines. Okay? So, now we have to solve this. This 4y minus 1 equal to 0. From here, we get y equal to 1 by 4. And if we get y equal to 1 by 4, put it here. So, this x minus 2y equal to 0. Now, x is equals to 2 times y. That is 2 into 1 by 4. That is nothing but x is equals to 1 by 2. So, we got the point. Means the chord of context will definitely pass through this point c, whose coordinates are 1 by 2 comma 1 by 4. So, we got the value of alpha as 1 by 2. And the value of beta as 1 by 4. Now it is asked to find the value of alpha square plus beta square. So, what will be our alpha square plus beta square? It will be 1 by 4. 1 by 4 plus 1 by 16. That is nothing but 4 plus 1. That is 5 by 16. Right? This will be our value of alpha square plus beta square. Okay. So, hope this exercise is clear to all of you. So, now this, we are done with this exercise. We will be back one soon with the next exercise that is exercise number 6. Okay. So, till then, take care.