Example 2. Find the equation of the circle that passes through the intersection points of the circles $\mathrm{x}^{2}+\mathrm{y}^{2}+6 \mathrm{y}-4=0$ and $x^{2}+y^{2}+6 y-28=0$, and whose center lies on the line $\mathrm{x}-\mathrm{y}-4=0$.
Solve the equation $\left(x^{2}+y^{2}+6 x-4\right)+\lambda\left(x^{2}+y^{2}+6 y-28\right)=0$ which represents the equation of a circle passing through the intersection points of two circles.

Rearranging gives $(1+\lambda) x^{2}+(1+\lambda) y^{2}+6 x$ $+6 \lambda y-4-28 \lambda=0$.
The center of the circle is $\left(-\frac{3}{1+\lambda} ;-\frac{3 \lambda}{1+\lambda}\right)$.
Since the center lies on the line $x-y-4=0$,
Therefore, $-\frac{3}{1+\lambda}+\frac{3 \lambda}{1+\dot{\lambda}}-4=0$.
Solving for $\lambda$ gives $\lambda=-7$. Thus,
$$
-6 x^{2}-6 y^{2}+6 x-42 y-4+196=0 .
$$

Rearranging gives $x^{2}+y^{2}-x+7 y-32=0$