The equation of the circle passing through $\left(1, 1\right)$ and through the points of intersection of the circles $x^2+y^2+13x-3y=0$ and $2x^2+2y^2+4x-7y-25=0$ is

Using family of circles, let the equation of the circle passing through the points of intersection of the circles $x^2+y^2+13x-3y=0$ and $2x^2+2y^2+4x-7y-25=0$ be $2x^2+2y^2+4x-7y-25+\lambda \left(x^2+y^2+13x-3y\right)=0$

Since it passes through $\left(1,1\right)$, so $2{\left(1\right)}^2+2{\left(1\right)}^2+4\left(1\right)-7\left(1\right)-25+\lambda \left({\left(1\right)}^2+{\left(1\right)}^2+13\left(1\right)-3\left(1\right)\right)=0$

$\Rightarrow \lambda =2$

So, the required equation of circle is $2x^2+2y^2+4x-7y-25+2\left(x^2+y^2+13x-3y\right)=0$

i.e. $4x^2+4y^2+30x-13y-25=0$