Example 13 Find the equation of the circle that is tangent to the circle $C: x^{2}+y^{2}-6 x-8 y+17=0$ at the point $(1,2)$ and has a radius of $\frac{5 \sqrt{2}}{2}$.
Consider the point $(1,2)$ as the point circle curve $\Gamma_{0}:(x-1)^{2}+(y-2)^{2}=0$. By property 9, the equation of the required circle is $x^{2}+y^{2}-6 x-8 y-17+\lambda\left[(x-1)^{2}+(y-2)^{2}\right]=0$.
That is, $\left(x-\frac{3 \lambda}{1+\lambda}\right)^{2}+\left(y-\frac{4+2 \lambda}{1+\lambda}\right)^{2}=\frac{8}{(1+\lambda)^{2}}$.
From $\frac{8}{(1+\lambda)^{2}}=\left(\frac{5 \sqrt{2}}{2}\right)^{2}$, we get $\lambda_{1}=-\frac{1}{5}$ or $\lambda_{2}=-\frac{9}{5}$.
Thus, we obtain $C_{1}:\left(x-\frac{7}{2}\right)^{2}+\left(y-\frac{9}{2}\right)^{2}=\left(\frac{5 \sqrt{2}}{2}\right)^{2}$ and $C_{2}:\left(x+\frac{3}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=\left(\frac{5 \sqrt{2}}{2}\right)^{2}$.
It is easy to see that the point $(1,2)$ satisfies the equation of circle $C_{1}$, and the distance between the centers of circle $C_{1}$ and circle $C$ equals the absolute value of the difference of their radii, so circle $C_{1}$ is internally tangent to circle $C$ at point $(1,2)$, and circle $C_{1}$ is the required one.
Similarly, circle $C_{1}$ is externally tangent to circle $C$ at point $(1,2)$, and circle $C_{2}$ is also the required one.