Four. (15 points) As shown in Figure 3, in the Cartesian coordinate system $x O y$, $\odot O: x^{2}+y^{2}=4$ intersects the positive half-axis of the $x$-axis at point $A$, $\odot A:(x-2)^{2}+y^{2}=r^{2}(r>0)$ intersects $\odot O$ at points $B$ and $C$.
(1) Find the minimum value of $\overrightarrow{A B} \cdot \overrightarrow{A C}$;
(2) Let $P$ be any point on $\odot O$ other than $B$ and $C$, and let the lines $P B$ and $P C$ intersect the $x$-axis at points $M$ and $N$, respectively. Find the maximum value of $S_{\triangle P O M} S_{\triangle P O N}$.
(1) By symmetry, without loss of generality, let point $B\left(x_{0}, y_{0}\right)$, $C\left(x_{0},-y_{0}\right)$. Then $x_{0}^{2}+y_{0}^{2}=4$.
Thus, $\overrightarrow{A B} \cdot \overrightarrow{A C}=\left(x_{0}-2\right)^{2}-y_{0}^{2}$
$$
=\left(x_{0}-2\right)^{2}-\left(4-x_{0}^{2}\right)=2\left(x_{0}-1\right)^{2}-2 \text {. }
$$

Since $-2<x_{0}<2$, when $x_{0}=1$, $\overrightarrow{A B} \cdot \overrightarrow{A C}$ achieves its minimum value of -2.
(2) Let point $P\left(x_{1}, y_{1}\right)\left(y_{1} \neq \pm y_{0}\right)$.
Then $x_{1}^{2}+y_{1}^{2}=4$,
$$
\begin{array}{l}
l_{P B}: y-y_{1}=\frac{y_{0}-y_{1}}{x_{0}-x_{1}}\left(x-x_{1}\right), \\
l_{P C}: y-y_{1}=\frac{-y_{0}-y_{1}}{x_{0}-x_{1}}\left(x-x_{1}\right) .
\end{array}
$$

By setting $y=0$, we get
$$
x_{M}=\frac{x_{1} y_{0}-x_{0} y_{1}}{y_{0}-y_{1}}, x_{N}=\frac{x_{1} y_{0}+x_{0} y_{1}}{y_{0}+y_{1}} \text {. }
$$

Then $x_{M} x_{N}=\frac{x_{1}^{2} y_{0}^{2}-x_{0}^{2} y_{1}^{2}}{y_{0}^{2}-y_{1}^{2}}$
$$
=\frac{\left(4-y_{1}^{2}\right) y_{0}^{2}-\left(4-y_{0}^{2}\right) y_{1}^{2}}{y_{0}^{2}-y_{1}^{2}}=4 \text {. }
$$

Therefore, $S_{\triangle P O M} S_{\triangle P O N}=\frac{1}{4}|O M||O N| y_{1}^{2}$
$$
=\frac{1}{4}\left|x_{M} x_{N}\right| y_{1}^{2}=y_{1}^{2} \text {. }
$$

Since $-2 \leqslant y_{1} \leqslant 2$, when $y_{1}=2$ or -2, $S_{\triangle P O M} S_{\triangle P O N}$ achieves its maximum value of 4.