 Hello and welcome to the session. In this session we will discuss a question which says that, find the focus of the point of intersection of two perpendicular tangents to the circle x square plus y square is equal to a square. Now before starting the solution of this question, we should know our result. And that is the equation of every tangent through the circle, square plus y square is equal to a square is y is equal to mx plus a into square root of 1 plus m square. Now this result will work out as a t-idea for solving out this question. And now we will start with the solution. Now here we have to find the focus of the point of intersection of two perpendicular tangents to the circle x square plus y square is equal to a square. Now let a t and a t dash be two perpendicular tangents through the circle with center o. And let the point of intersection of a t and a t dash be x from y 1. Now using the result which is given in the key idea, the equation every tangent through the circle, square plus y square is equal to a square is y is equal to mx plus a into square root of 1 plus m square. Now if this tangent passes through the point that is this point, which is a x-ray ribbon then now here says x-ray and y as ribbon as this tangent is passing through this point. So we are getting y1 is equal to mx1 plus a into square root of 1 plus m square. This implies y1 minus mx1 is equal to a into square root of 1 plus m square. Further this implies on square both sides y1 minus mx1 whole square is equal to a square into 1 plus m square the whole. This implies y1 square plus m square x1 square minus 2 y1 mx1 is equal to a square plus a square m square. Which further implies y1 square plus m square x1 square minus 2 y1 mx1 minus a square minus a square m square is equal to 0. Now this implies on combining these terms we get m square into x1 square minus a square the whole minus 2 into x1 into y1 into m plus within brackets y1 square minus a square is equal to 0. Now this is a quadratic equation in m and m2 is equation b1 and m2 are the roots of 1. Therefore 2 over a which implies m1 into m2 is equal to c which is this that is y1 square minus a square over that is this that is x1 square minus a square. Now let us name this as 2. Now suppose on the quadratic equation the slopes a t and a t dash to each other a t and a t dash are perpendicular each other. Therefore product of the slopes is equal to minus 1. Now let us name this as this is the equation number 2 and this is the equation number 3 from t and 3 y1 square minus a square whole upon x1 square minus a square is equal to minus 1. Which further implies y1 square minus a square is equal to minus x1 square plus a square plus y1 square is equal to 2a square. Therefore the layer cost x1 y1 that is the point of intersection of the perpendicular tangents to the circle is plus y1 square is equal to 2a square. So this is the solution of the given question and that is all for this session. Hope you all have enjoyed this session.