Example 7 Let $P$ and $Q$ be points on the parabola $y^{2}=x$ and the circle $(x-3)^{2}+y^{2}=1$ respectively, find $|P Q|_{\operatorname{min}}$.


Solve As shown in Figure 11-4, let the center of the circle be $C(3,0)$, with a radius of 1. If $|P Q|$ is minimized, then $|C P|=|P Q|+|C Q|$ is also minimized, so $C, P, Q$ are collinear. This reduces the problem to finding a point $P$ on the parabola such that its distance to point $C$ is minimized. For this, let $P\left(y_{0}^{2}, y_{0}\right)$, then $\left\lvert\, P\left(\left.\right|^{2}=\left(y_{0}^{2}-3\right)^{2}+y_{0}^{2}=\left(y_{0}^{2}-\frac{5}{2}\right)^{2}+\frac{11}{4} \geqslant \frac{11}{4}\right.\right.$, so the minimum value of $|P Q|$ is $\frac{\sqrt{11}}{2}-1$.