9. (16 points) Given $\odot O: x^{2}+y^{2}=4$, circle $M$ :
$$
(x-5 \cos \theta)^{2}+(y-5 \sin \theta)^{2}=1(\theta \in \mathbf{R}) \text {, }
$$

Through any point $P$ on circle $M$, draw two tangents $P E$ and $P F$ to $\odot O$, with the points of tangency being $E$ and $F$. Try to find the minimum value of $\overrightarrow{P E} \cdot \overrightarrow{P F}$.
9. The center of circle $M$ is on the circle $x^{2}+y^{2}=25$.
Let $|P E|=|P F|=d$.
In the right triangle $\triangle P E O$, it is easy to see that
$4 \leqslant|P O| \leqslant 6,|O E|=2$.
Thus, $2 \sqrt{3} \leqslant d \leqslant 4 \sqrt{2}$.
Also, $\overrightarrow{P E} \cdot \overrightarrow{P F}=|\overrightarrow{P E}||\overrightarrow{P F}| \cos \angle E P F$
$=d^{2} \cos \angle E P F$.
Let $\angle O P E=\theta$. Then,
$\tan \theta=\frac{|O E|}{|P E|}=\frac{2}{d}$.
Thus, $\cos \angle E P F=\cos 2 \theta$
$=\cos ^{2} \theta-\sin ^{2} \theta$
$=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\frac{d^{2}-4}{d^{2}+4}$.
Therefore, $\overrightarrow{P E} \cdot \overrightarrow{P F}=d^{2} \cdot \frac{d^{2}-4}{d^{2}+4}$
$=\left(d^{2}+4\right)+\frac{32}{d^{2}+4}-12$.
Let $d^{2}+4=x$. Then $x \in[16,36]$.
Using the function $f(x)=x+\frac{32}{x}$ which is monotonically increasing in the interval $[16,36]$, we know that when $d^{2}+4=16$,
$$
(\overrightarrow{P E} \cdot \overrightarrow{P F})_{\min }=6
$$