 Hello, and welcome to the session. In this session, I discuss a question which says that find location of the common part of the two circles, x squared plus y square minus 4 x minus 10 y minus 7 is equal to 0. And 2x square plus 2y square minus pi x plus 3 y plus 2 is equal to 0. Show that this part is perpendicular to the line joining the centers of the two circles. Now, before starting the solution of this question, we should know our result. Let s1 is equal to 0, and s2 is equal to 0, where s1 is x square plus y square plus 2g1x plus 2f1y plus cy. And s2 is x square plus y square plus 2g2x plus 2f2y plus c2 are the equations of two circles, constant, then, into s2 is equal to 0, where here it is not equal to minus 1, represents two of circles passing through the points of intersection of the given circles. If s1 is equal to minus 1, then we get s1 minus s2 is equal to 0. That is, what in here is equal to minus 1 here. And this is the equation of this field line, which passes intersection of the given circles if we have any. Now, this result will work out as a t-idea for solving out this question. And now, we will start with the solution. Here, the equations of the circles are given to us. The equations of the two given circles are y square minus 4x minus 10y minus 7 is equal to 0. s2 y square minus 5x plus 3y plus 2 is equal to 0. Or you can write s2x square plus y square minus 5y2x plus 3y2y plus 1 is equal to 0. Now, we have to find the equation of the common part of the two circles. Now, using this result, which is given in the key-idea, the equation of the common part of the two circles is given by is equal to 0. And this is s1 and this is s2. Now, putting the number of s1 and s2 here, this implies plus y square minus 4x minus 10y minus 7 minus of x square plus y square minus 5y2x plus 3y2y plus 1 the whole is equal to 0. This implies plus y square minus 4x minus 10y minus 7 minus x square minus y square plus 5y2x minus 3y2y minus 1 is equal to 0. Now, these terms will be cancelled with each other. So this implies minus 4x plus 5y2x plus 2y minus 7 minus 1 is equal to 0. So solving this implies minus 3y2x, which therefore implies minus 3y plus 23y plus 16 is equal to 0. It's perpendicular to the line joining the centers of the two circles, plus 1. And this is the circle, s is the general equation of the circle, whose coordinates of center are given by the circle s1. So the centers in s10 by 2 to c1, the center c2, will be given by minus 4 and minus 3 by 4. Now, let these points 1 and c2 minus y1 over x2 minus m is equal to minus 3 by 4, 3 by 4, which is further equal to 23 by 3, c1, c2 is equal to 23 by 3. Now, let this m1, which is equal to 23 by 3, is equal to minus coefficient of x or coefficient of y, minus 3 by 23. Now, it will be equal to 2 by 3 into minus 3 by 23 into m2 is equal to minus 1. Now, if I prepare to clear the centers c1 given question, and that's all for this.