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

The equation of the required circle is

$x^2+y^2+13x-3y+\lambda \left(x^2+y^2+2x-\frac{7}{2}y-\frac{25}{2}\right)=0$

This passes through (1, 1). 

$\therefore 12+\lambda (-12)=0\Rightarrow \lambda =1.$

Hence, the required circle is

$\begin{array}{l} x^2+y^2+13x-3y+x^2+y^2+2x-\frac{7}{2}y-\frac{25}{2}=0\\ \Rightarrow 4x^2+4y^2+30x-13y-25=0\end{array}$