Five. (15 points) As shown in Figure 4, $AOB$ is a quarter of a unit circle. The center of the semicircle $O_{1}$ is on $OA$ and is internally tangent to the arc $\overparen{AB}$ at point $A$. The center of the semicircle $O_{2}$ is on $OB$ and is internally tangent to the arc $\overparen{AB}$ at point $B$. The semicircle $O_{1}$ is tangent to the semicircle $O_{2}$. Let the sum of the radii of the two semicircles be $x$, and the sum of their areas be $y$.
(1) Try to establish the analytical expression of the function $y$ with $x$ as the independent variable;
(2) Find the minimum value of the function $y$.


(1) Let the radius of $\odot C_{1}$ be $R$, and the radius of $\mathrm{S} \mathrm{O}_{2}$ be $r$ (as shown in Figure 6).
$$
\begin{aligned}
y= & \frac{1}{2} \pi\left(R^{2}+r^{2}\right) \\
= & \frac{1}{2} \pi\left[(R+r)^{2}\right. \\
& -2 R r] .
\end{aligned}
$$

Connecting $O_{1} O_{2}$, this line must pass through the point of tangency of the two circles.
From the right triangle $\triangle O_{1} O O_{2}$, we know
$$
(R+r)^{2}=(1-R)^{2}+(1-r)^{2} \text {, }
$$

which simplifies to
$$
\begin{array}{l}
R r+R+r=1 . \\
\begin{aligned}
\therefore y & =\frac{1}{2} \pi\left\{(R+r)^{2}-2[1-(R+r)]\right\} \\
& =\frac{\pi}{2}\left[(R+r)^{2}+2(R+r)-2\right] .
\end{aligned}
\end{array}
$$

Given $R+r=x$, then $y=\frac{\pi}{2}\left(x^{2}+2 x-2\right)$.
$$
\begin{array}{l}
\text { (2) } \because R+r \geqslant 2 \sqrt{R r}, \\
\therefore \frac{(R+r)^{2}}{4} \geqslant R r . \\
\text { Also, } \because R r=1-(R+r), \\
\therefore(R+r)^{2}+4(R+r)-4 \geqslant 0 . \\
\because R+r>0, \therefore R+r \geqslant 2(\sqrt{2}-1),
\end{array}
$$

which means $x \geqslant 2(\sqrt{2}-1)$.
Thus, the function $y=\frac{\pi}{2}\left(x^{2}+2 x-2\right)$ has its minimum value when $x=2(\sqrt{2}-1)$, which is $y_{\text {min }}=(3-2 \sqrt{2}) \pi$.
(Provided by Ren Chaoyu, Taiyuan Education Institute)