Question

Show that the equation of the circle passing through $(1,1)$ and the points of intersection of the circles $x^2+y^2+13x-13y=0$ and $2x^2+2y^2+4x-7y-25=0$ is $4x^2+4y^2+30x-13y-25=0$.

Answer 1

$S_1:x^2+y^2+13x-13y=0$

$S_2:2x^2+2y^2+4x-7y-25=0$

Equation of family of circles passing through points of intersection of $S_1$ & $S_2$ is

$S_2+\lambda S_1=0$

$\Rightarrow x^2(2+\lambda )+y^2(2+\lambda )+x(13\lambda +4)-y(13\lambda +7)-25=0$

Satisfying $(1,1)$

$2+\lambda +2+\lambda +13\lambda +4-13\lambda -7-25=0$

$\Rightarrow \lambda =12$

$\Rightarrow$ circle is $14x^2+14y^2+160x-163y-25=0$