Example 2 Given $x^{2}+y^{2}=a^{2}, a^{2}<1$. Try to find: $د=\sqrt{1-x^{2}}+\sqrt{1-y^{2}}$ the maximum and minimum values. (1993, Shandong Province Junior High School Mathematics Competition)
$$
\begin{aligned}
\text { Solution: } s & =\sqrt{1-x^{2}}+\sqrt{1-y^{2}} \\
& =\sqrt{\left(\sqrt{1-x^{2}}+\sqrt{1-y^{2}}\right)^{2}} \\
& =\sqrt{2-\left(x^{2}+y^{2}\right)+2 \sqrt{1-\left(x^{2}+y^{2}\right)+x^{2} y^{2}}} . \\
\because x^{2} & +y^{2}=a^{2}, \\
\therefore s & =\sqrt{2-a^{2}+2 \sqrt{1-a^{2}+a^{2} x^{2}-x^{4}}} \\
& =\sqrt{2-a^{2}+2 \sqrt{1-a^{2}+\frac{a^{4}}{4}-\left(\frac{a^{2}}{2}-x^{2}\right)^{2}} .}
\end{aligned}
$$

To find the maximum and minimum values of $s$, we only need to consider the minimum and maximum values of $\left(\frac{a^{2}}{2}-x^{2}\right)^{2}$.

When $\frac{a^{2}}{2}-x^{2}=0$, i.e., $x^{2}=\frac{a^{2}}{2}$, then $y^{2}=\frac{a^{2}}{2}$. At this time,
$s_{\text {max }}=\sqrt{1-\frac{a^{2}}{2}}+\sqrt{1-\frac{a^{2}}{2}}=\sqrt{4-2 a^{2}}$.
When $\left(\frac{a^{2}}{2}-x^{2}\right)^{2}=\left(\frac{a}{2}\right)^{2}$, i.e., $x^{2}=0$, then $y^{2}=a^{2}$. At this time,
$$
s_{\text {min }}=1+\sqrt{1-a^{2}} .
$$
