Example 4. Let $z=\frac{x^{2}}{4}-y^{2}$, when $x^{2}+y^{2} \leqslant 4$, find $z_{max}$ and $z_{min}$.
Solve: First, make the domain of $x, y$ (as shown in Figure 4), which is a closed circular region centered at $O$ with a radius of 2.
$$
\text { Let } \frac{x^{2}}{4}-y^{2}=k \text {. }
$$
(i) When $\mathrm{k}=0$
$$
\begin{array}{l}
\text { then, } \frac{x^{2}}{4}-y^{2}=0 \Leftrightarrow \\
\left(\frac{x}{2}\right)^{2}-y^{2}=0 \Longleftrightarrow \\
\left(\frac{x}{2}+y\right)\left(\frac{x}{2}-y\right) \\
=0 .
\end{array}
$$

It represents two intersecting lines passing through point $\mathrm{O}$, meaning that the function values corresponding to the points on these two lines are all 0.
(ii) When $k>0$,
$$
\frac{x^{2}}{4}-y^{2}=k \Leftrightarrow-\frac{x^{2}}{(2 \sqrt{k})^{2}}-\frac{y^{2}}{\sqrt{k}}=1 .
$$

It represents a family of hyperbolas centered at $\mathrm{O}$, with the real axis on the $\mathrm{x}$-axis and the asymptotes being the two lines $\mathrm{y}=\frac{\mathrm{x}}{2}$ and $\mathrm{y}=-\frac{\mathrm{x}}{2}$ when $\mathrm{k}=0$. When such a family of hyperbolas passes through region $D$, the maximum square of the imaginary semi-axis is the maximum positive value of the function in $D$. The hyperbola in the family with vertices at $A(2,0)$ and $B(-2,0)$ has the largest imaginary semi-axis of 1, $\therefore:_{\max}=1^{2}=1$, or $\max =\frac{2^{2}}{4}-0^{2}=1$.
(iii) When $k<0$,
$$
\frac{x^{2}}{4}-y^{2}=k \Leftrightarrow \frac{-x^{2}}{(2 \sqrt{|k|})^{2}}+\frac{y^{2}}{(\sqrt{|k|})^{2}}=1 .
$$

Similarly, we can get $\approx_{\min}=-4$, or $\min =\frac{0}{4}-2^{2}=-4$.