9. (16 points) In the plane region
$$
\left\{\begin{array}{l}
x-\sqrt{3} y+4 \geqslant 0, \\
x+\sqrt{3} y+4 \geqslant 0, \\
x \leqslant 2
\end{array}\right.
$$

draw a circle, and denote the largest circle within this region as $\odot M$.
(1) Find the equation of $\odot M$;
(2) $\odot M$ intersects the $x$-axis at points $A$ and $B$. For a moving point $P$ inside the circle such that $|P A|,|P M|,|P B|$ form a geometric sequence, find the range of $\overrightarrow{P A} \cdot \overrightarrow{P B}$.
9. (1) (Method 1) From the problem, we know that $\odot M$ is the incircle of the triangular region.

Let the equation of $\odot M$ be $(x-a)^{2}+(y-b)^{2}=r^{2}(r>0)$, then the point $(a, b)$ is inside the given region. Therefore,

we solve to get $a=0, b=0, r=2$, so the equation of the required circle is $x^{2}+y^{2}=4$.
(Method 2) From the problem, we know that $\odot M$ is the incircle of the triangular region.
Let the region determined by the equations in the problem be $\triangle D E F$ (as shown in Figure 4). Since the line $x-\sqrt{3} y+4=0$ is symmetric to the line $x+\sqrt{3} y+4=0$ about the $x$-axis, and the inclination angle of $x-\sqrt{3} y+4=0$ is $30^{\circ}$, one of the internal angles of the triangle is $60^{\circ}$.
The line $x=2$ is perpendicular to the angle bisector of $\angle E D F$, with point $H(2,0), D(-4,0)$, so $\triangle D E F$ is an equilateral triangle, and the height of the triangle is 6. The center of the incircle is the centroid of $\triangle D E F$, i.e., $O(0,0)$, and the radius is $|O H|=2$, so the equation of the required circle is $x^{2}+y^{2}=4$.
(2) Without loss of generality, let $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right), x_{1}<x_{2}$. From $x^{2}=4$, we get $A(-2,0), B(2,0)$. Let $P(x, y)$, and since $|P A|,|P B|,|P C|$ form a geometric sequence, we have
$$
\sqrt{(x+2)^{2}+y^{2}} \cdot \sqrt{(x-2)^{2}+y^{2}}=x^{2}+y^{2} \Rightarrow x^{2}-y^{2}=2 .
$$

Thus,
$$
\begin{aligned}
\overrightarrow{P A} \cdot \overrightarrow{P B} & =(-2-x,-y) \cdot(2-x,-y)=x^{2}-4+y^{2} \\
& =2\left(y^{2}-1\right) .
\end{aligned}
$$

Since point $P$ is inside $\odot M$, we have
$$
\left\{\begin{array}{l}
x^{2}+y^{2}<4, \\
x^{2}-y^{2}=2,
\end{array}\right.
$$

From this, we get $y^{2}<1$. Therefore, the range of $\overrightarrow{P A} \cdot \overrightarrow{P B}$ is $[-2,0)$.