Example 3. Given $\odot \mathrm{O}_{1}: \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{x}-4 \mathrm{y}+1$ $=0, \odot \mathrm{O}_{2}: \mathrm{x}^{2}+\mathrm{y}^{2}-10 \mathrm{x}-4 \mathrm{y}-71=0$. If a moving circle is externally tangent to $\odot \mathrm{O}_{1}$ and internally tangent to $\odot \mathrm{O}_{2}$, find the equation of the locus of the center of the moving circle.
$$
\begin{array}{c}
\text { Solving } \odot \mathrm{O}_{1}: \quad(\mathrm{x} \\
+1)^{2}+(\mathrm{y}-2)^{2}=4, \\
\text { radius } \mathrm{R}_{1}=2 ; \\
\odot \mathrm{O}_{2}: \\
(\mathrm{x}-5)^{2}+(\mathrm{y}-2)^{2} \\
=100, \text { radius } \mathrm{R}_{2}=10 .
\end{array}
$$

Let the center of the moving circle be $P(x, y)$, and its radius be $R$. As shown in Figure 4, $| \mathrm{PO}_{1}|=\mathrm{R}+\mathrm{R}_{1}, | \mathrm{PO}_{2} |=\mathrm{R}_{2}-\mathrm{R}$. Therefore, $| \mathrm{PO}_{1}|+| \mathrm{PO}_{2}|=R_{1}+R_{2}=12$.

It is evident that the trajectory of the moving circle's center is an ellipse. It is not difficult to find the equation of the trajectory of the moving circle's center.
$$
\frac{(x-2)^{2}}{36}+\frac{(y-2)^{2}}{27}=1 \text {. }
$$