Five, let $x^{2}+y^{2}=a^{2}, a^{2}<1$. Try to find the maximum and minimum values of $S=\sqrt{1-x^{2}}+\sqrt{1-y^{2}}$.
$$
\begin{aligned}
& \text { V. Solution } S=\sqrt{1-x^{2}}+\sqrt{1-y^{2}} \\
= & \sqrt{\left(\sqrt{1-x^{2}}+\sqrt{1-y^{2}}\right)^{2}} \\
= & \sqrt{2-\left(x^{2}+y^{2}\right)+2 \sqrt{1-\left(x^{2}+y^{2}\right)+x^{2} y^{2}}} \\
= & \sqrt{2-a^{2}+2 \sqrt{1-a^{2}}+x^{2}\left(a^{2}-x^{2}\right)} \\
= & \sqrt{2-a^{2}+2 \sqrt{1-a^{2}+\frac{a^{4}}{4}-\left(x^{2}-\frac{a^{2}}{2}\right)^{2}}} .
\end{aligned}
$$

From the above equation, we can see that:
(1) When $x=0$,
$$
S_{\text {max }}=1+\sqrt{1-a^{2}} \text {. }
$$
(2) When $x^{2}=\frac{a^{2}}{2}$,
$$
\begin{aligned}
S_{\text {min }} & =\sqrt{1-\frac{a^{2}}{2}}+\sqrt{1-\frac{a^{2}}{2}} \\
& =\sqrt{4-2 a^{2}} .
\end{aligned}
$$
(Provided by Nawei Liang)
$$