Example 4. Given $(x-2)^{2}+(y-2)^{2}=1$, find the extremum of $x \cdot y$.


Solve: The graph of $(x-2)^{2}+(y-2)^{2}=1$ is a circle with center at point $(2,2)$ and radius 1. This circle is symmetrical about the line $y=x$, so the hyperbola family $y=\frac{c}{x}$ that intersects the circle $(x-2)^{2}+(y-2)^{2}=1$ has extremal values of the coefficient at the hyperbolas passing through points $A$ and $B$. Points $A$ and $B$ are the intersection points of the line $y=x$ and the circle $(x-2)^{2}+(y-2)^{2}=1$.
From the graph, we get $O A=2 \sqrt{2}-1$,
$$
O B=2 \sqrt{2}+1 \text {. }
$$

Thus, the coordinates of point $A$ are $\left(\frac{4-\sqrt{2}}{2}\right.$, $\left.\frac{4-\sqrt{2}}{2}\right)$, and the coordinates of point $B$ are $\left(\frac{4+\sqrt{2}}{2}\right.$, $\left.\frac{4+\sqrt{2}}{2}\right)$.
Therefore, the minimum value of $m \cdot y$ is
$$
4-\frac{v}{2} \cdot \frac{4-\sqrt{2}}{2}=\frac{9-4 \sqrt{2}}{2} \text {, }
$$
and the maximum value of $x \cdot y$ is
$$
\frac{4+\sqrt{2}}{2} \cdot \frac{4+\sqrt{2}}{2}=\frac{9+4 \sqrt{2}}{2} \text {. }
$$