The equation of a circle passing through the points of intersection of the circles $x^2+y^2-4x-6y-12=0,x^2+y^2+6x+4y-12=0$ and having radius $\sqrt{13}$ is

The equation of a circle passing through the points of
intersection of the circles,

$\left(x^2+y^2-4x-6x-12\right)+\lambda \left(x^2+y^2+6x+4y-12\right)=0$

$x^2(1+\lambda )+y^2(1+\lambda )+x(6\lambda -4)+y(4\lambda -6)-12\lambda -12=0$

$\text{Rewrite the above equation. }$

$x^2+y^2+\frac{x(6\lambda -4)}{(1+\lambda )}+\frac{y(4\lambda -6)}{(1+\lambda )}-12=0$

$\text{On comparing in equations with }$$x^2+y^2+2gx+2fy+c=0$

$g=\frac{(3\lambda -2)}{(1+\lambda )},f=\frac{(2\lambda -3)}{(1+\lambda )},c=-12$

$\text{It is given that the radius is }$$\sqrt{13}$

$13=\frac{(3\lambda -2{)}^2}{(1+\lambda {)}^2}+\frac{(2\lambda -3{)}^2}{(1+\lambda {)}^2}+12$

$13(1+\lambda {)}^2=(3\lambda -2{)}^2+(2\lambda -3{)}^2+12(1+\lambda {)}^2$

$\text{After simplification, }$

$13{\lambda }^2+13+26\lambda =25{\lambda }^2+25$

$12{\lambda }^2-26\lambda +12=0$

$6{\lambda }^2-13\lambda +6=0$

$(2\lambda -3)(3\lambda -2)=0$

$\lambda =\frac{3}{2},\frac{2}{3}$

$\text{The required equation when }$$\lambda =\frac{3}{2}$

$x^2\left(1+\frac{3}{2}\right)+y^2\left(1+\frac{3}{2}\right)+x\left(6\times \frac{3}{2}-4\right)+y\left(4\times \frac{3}{2}-6\right)-12\times \frac{3}{2}-12=0$

$\text{Further simplification, }$

$x^2+y^2+2x-12=0$

$\text{The required equation when }$ $\lambda =\frac{2}{3}$

$x^2\left(1+\frac{2}{3}\right)+y^2\left(1+\frac{2}{3}\right)+x\left(6\times \frac{2}{3}-4\right)+y\left(4\times \frac{2}{3}-6\right)-12\times \frac{2}{3}-12=0$

$\text{Further simplification, }$

$x^2+y^2-2x-12=0$