Example $8 . \odot O_{1}$ and $\odot O_{2}$ intersect at $P$ and $Q$, a line through $P$ intersects $\odot O_{1}$ at $A$ and $\odot O_{2}$ at $B$, such that $P A \cdot P B$ is maximized.
Analysis: As shown in the figure,
$$
\begin{array}{l}
P A=2 r \sin \alpha, P B=2 R \sin \beta, \\
\therefore P A \cdot P B=4 R r \sin \alpha \sin \beta \\
=2 R r[\cos (\alpha-\beta)-\cos (\alpha+\beta)] .
\end{array}
$$
Since $\angle A$ and $\angle B$ are constants, then $\alpha+\beta$ is a constant.
Therefore, when $\cos (\alpha-\beta)=1$, $P A \cdot P B$ is maximized.
At this time, $\alpha=\beta$. Draw $O_{1} M \perp P A, O_{2} N \perp P B$, and extend $\mathrm{O}_{2} P$, then
$$
\begin{array}{l}
\angle M O_{1} P=\alpha=\beta=\angle N O_{2} P . \\
\therefore \angle 1=\angle 2=\angle 3 .
\end{array}
$$

Therefore, $A P$ should be the external angle bisector of $\triangle O_{1} O_{2} P$. The construction method is omitted.