Example 5 As shown in Figure 1, it is known that point $P$ moves on the circle $x^{2}+(y-4)^{2}=1$, and point $Q$ moves on the ellipse $\frac{x^{2}}{9}+y^{2}=1$. Try to find the maximum value of $|PQ|$.
(1994, Sichuan Province High School Mathematics Competition)
Analysis: First, fix point $Q$ on the ellipse. It is evident that $|P Q|$ is maximized when $P Q$ passes through the center $O_{1}$. Therefore, to find the maximum value of $|P Q|$, we only need to find the maximum value of $\left|O_{1} Q\right|$.
Let $Q(x, y)$, then
$$
\left|O_{1} Q\right|^{2}=x^{2}+(y-4)^{2} \text {. }
$$

Since $Q$ is on the ellipse, we have $\frac{x^{2}}{9}+y^{2}=1$, which means
$$
x^{2}=9\left(1-y^{2}\right) \text {. }
$$

Substituting equation (2) into equation (1) yields,
$$
\begin{array}{l}
\left|O_{1} Q\right|^{2}=9\left(1-y^{2}\right)+(y-4)^{2} \\
=-8 y^{2}-8 y+25=-8\left(y+\frac{1}{2}\right)^{2}+27 .
\end{array}
$$

Since point $Q$ moves on the ellipse, $-1 \leqslant y \leqslant 1$. Therefore, when $y=-\frac{1}{2}$,
$$
\left|O_{1} Q\right|_{\max }=3 \sqrt{3},|P Q|_{\max }=3 \sqrt{3}+1 \text {. }
$$

Note: In problems involving the analytic geometry of circles, it is often necessary to flexibly apply the relevant properties of circles.