Example 1. Given circles
$$
\begin{array}{l}
C_{1}:(x-a)^{2}+y^{2} \\
=a^{2} \text { and } C_{2}:(x- \\
2 a)^{2}+y^{2}=4 a^{2}
\end{array}
$$
$(a>0)$. Find the locus of the centers of circles that are externally tangent to $C_{1}$ and internally tangent to $C_{2}$.
Let $\mathrm{P}(\mathrm{x}, \mathrm{y})$ be the center of the moving circle (positioning element)
(Figure 6), and the points of tangency be $\mathrm{K}$ and $\mathrm{H}$. Then, $\mathrm{PH} = \mathrm{PK}$. But $\mathrm{PH} = \sqrt{(x-a)^{2}+y^{2}} - a$, $\mathrm{PK} = 2a - \sqrt{(x-2a)^{2}+y^{2}}$. Substituting these into the previous equation and simplifying, we get
$$
8 x^{2} + 9 y^{2} - 24 a x = 0
$$

This is the desired equation.