Example 5. Given that the center of circle $C$ lies on $x-y-4=0$ and it passes through the intersection points of two circles $C_{1}: x^{2}+y^{2}-4 x-3=0$, $C_{2}: x^{2}+y^{2}-4 y-3=0$. Find the equation of the circle.
The equation of the family of circles passing through the intersection points of circles $C_{1}, C_{2}$ is
$$
\begin{aligned}
& x^{2}+y^{2}-4 x-3 \\
+ & \lambda\left(x^{2}+y^{2}-4 y-3\right) \\
= & 0 .
\end{aligned}
$$

That is,
$$
\begin{array}{l}
\quad(1+\lambda) x^{2}+(1 \\
+\lambda) y^{\prime 2}-4 x-4 \lambda y \\
-3(1+\lambda)=0 .
\end{array}
$$

When $\lambda \neq -1$, the coordinates of the center of the circle are
$$
\left(\frac{2}{1+\lambda}, \frac{2 \lambda}{1+\lambda}\right) .
$$

The center of the circle lies on the line $x-y-4=0$. Therefore, we have
$$
\frac{2}{1+\lambda}-\frac{2 \lambda}{1+\lambda}=4 \text {. }
$$

Solving this, we get
$$
\lambda=-\frac{1}{3} .
$$

The equation of the circle is
$$
x^{2}+y^{2}-6 x+2 y-3=0 \text {. }
$$

Using the equations of the family of circles and the family of lines to solve problems, it is only necessary to determine one constant, greatly simplifying the solution process and improving efficiency.