Example 8 (1) As shown in Figure 8, within the square $ABCD$, there are two moving circles $\odot O_{1}$ and $\odot O_{2}$ that are externally tangent to each other. $\odot O_{1}$ is tangent to sides $AB$ and $AD$, and $\odot O_{2}$ is tangent to sides $BC$ and $CD$. If the side length of the square $ABCD$ is $1$, and the radii of $\odot O_{1}$ and $\odot O_{2}$ are $r_{1}$ and $r_{2}$ respectively, find:
(i) the relationship between $r_{1}$ and $r_{2}$;
(ii) the minimum value of the sum of the areas of $\odot O_{1}$ and $\odot O_{2}$.
(2) As shown in Figure 9, if the square $ABCD$ is changed to a rectangle with a width of 1 and a length of $\frac{3}{2}$, and all other conditions remain the same, does the sum of the areas of $\odot O_{1}$ and $\odot O_{2}$ have a minimum value? If not, explain why; if so, find this minimum value. ${ }^{[8]}$
(2010, National Junior High School Mathematics Competition, Tianjin Preliminary Contest)
(1) (i) From the given conditions, points $O_{1}$ and $O_{2}$ are both on the diagonal $AC$ of the square.
By symmetry,
$$
\begin{array}{l}
\sqrt{2} r_{1} + r_{1} + r_{2} + \sqrt{2} r_{2} = AC = \sqrt{2} \\
\Rightarrow r_{1} + r_{2} = 2 - \sqrt{2}.
\end{array}
$$
(ii) From $r_{2} = 2 - \sqrt{2} - r_{1} \leqslant \frac{1}{2}$, we have
$$
\begin{array}{l}
\frac{3}{2} - \sqrt{2} \leqslant r_{1} \leqslant \frac{1}{2}. \\
\text{Thus, } S = \pi (r_{1}^{2} + r_{2}^{2}) \\
= \pi \left[r_{1}^{2} + (2 - \sqrt{2} - r_{1})^{2}\right] \\
= \pi \left[2 r_{1}^{2} - 2(2 - \sqrt{2}) r_{1} + 6 - 4 \sqrt{2}\right] \\
= \pi \left[2 \left(r_{1} - \frac{2 - \sqrt{2}}{2}\right)^{2} + 3 - 2 \sqrt{2}\right].
\end{array}
$$

Since $\frac{3}{2} - \sqrt{2} \leqslant \frac{2 - \sqrt{2}}{2} \leqslant \frac{1}{2}$, when $r_{1} = \frac{2 - \sqrt{2}}{2}$,
$$
S_{\text{min}} = (3 - 2 \sqrt{2}) \pi.
$$
(2) As shown in Figure 10.

In the right triangle $\triangle O_{1} O_{2} K$,
$$
\begin{array}{l}
O_{1} O_{2}^{2} = K O_{1}^{2} + K O_{2}^{2} \\
\Rightarrow (r_{1} + r_{2})^{2} = \left(\frac{3}{2} - r_{1} - r_{2}\right)^{2} + \left(1 - r_{1} - r_{2}\right)^{2} \\
\Rightarrow (r_{1} + r_{2})^{2} - 5(r_{1} + r_{2}) + \frac{13}{4} = 0 \\
\Rightarrow r_{1} + r_{2} = \frac{5}{2} \pm \sqrt{3}.
\end{array}
$$

Since $r_{1} + r_{2} < 1 + \frac{3}{2}$, we have
$$
r_{1} + r_{2} = \frac{5}{2} - \sqrt{3}.
$$

Thus, $S = \pi (r_{1}^{2} + r_{2}^{2}) \geqslant \pi \frac{(r_{1} + r_{2})^{2}}{2}$
$$
= \pi \frac{\left(\frac{5}{2} - \sqrt{3}\right)^{2}}{2} = \left(\frac{37}{8} - \frac{5 \sqrt{3}}{2}\right) \pi.
$$