Example 11 (1994 Sichuan Province High School Competition Question) Given that point $P$ moves on the circle $x^{2}+(y-4)^{2}=1$, and point $Q$ moves on the ellipse $\frac{x^{2}}{9}+y^{2}=1$, find the maximum value of $|P Q|$.

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Solving: Since the center is $O(0,4), Q(3 \cos \theta, \sin \theta)$, thus
$$
\begin{aligned}
|O Q|^{2} & =(3 \cos \theta)^{2}+(\sin \theta-4)^{2}=9 \cos ^{2} \theta+\sin ^{2} \theta-8 \sin \theta+16 \\
& =9-9 \sin ^{2} \theta+\sin ^{2} \theta-8 \sin \theta+16=27-8\left(\sin ^{2} \theta+\sin \theta+\frac{1}{4}\right) \\
& =27-8\left(\sin \theta+\frac{1}{2}\right)^{2} \leqslant 27 .
\end{aligned}
$$

When $\theta=-\frac{\pi}{6}$, the equality holds, so $|O Q| \leqslant \sqrt{27} .|P Q| \leqslant|P O|+|O Q|=1+\sqrt{27}=1+3 \sqrt{3}$, when $\theta=-\frac{\pi}{6}$, the equality holds, hence the maximum value of $|P Q|$ is $3 \sqrt{3}+1$.