9. Given the hyperbola $x^{2}-y^{2}=2$ with left and right foci $F_{1}, F_{2}$, a moving line through point $F_{2}$ intersects the hyperbola at points $A, B$.
(1) If a moving point $M$ satisfies $\overrightarrow{F_{1} M}=\overrightarrow{F_{1} A}+\overrightarrow{F_{1} B}+\overrightarrow{F_{1} O}$ (where $O$ is the origin), find the equation of the trajectory of point $M$;
(2) Does there exist a fixed point $C$ on the $x$-axis such that $\overrightarrow{C A} \cdot \overrightarrow{C B}$ is a constant? If it exists, find the coordinates of point $C$; if not, explain the reason.
9. Solution: Given $F_{1}(-2,0), F_{2}(2,0)$, let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
(1) Let $M(x, y)$, then $\overrightarrow{F_{1} M}=(x+2, y), \overrightarrow{F_{1} A}=\left(x_{1}+2, y_{1}\right)$,
$\overrightarrow{F_{1} B}=\left(x_{2}+2, y_{2}\right), \overrightarrow{F_{1} O}=(2,0)$, from $\overrightarrow{F_{1} M}=\overrightarrow{F_{1} A}+\overrightarrow{F_{1} B}+\overrightarrow{F_{1} O}$ we get
$\left\{\begin{array}{l}x+2=x_{1}+x_{2}+6, \\ y=y_{1}+y_{2}\end{array}\right.$ which is $\left\{\begin{array}{l}x_{1}+x_{2}=x-4, \\ y_{1}+y_{2}=y\end{array}\right.$, thus the midpoint coordinates of $A B$ are $\left(\frac{x-4}{2}, \frac{y}{2}\right)$.
When $A B$ is not perpendicular to the $x$-axis, $\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{\frac{y}{2}}{\frac{x-4}{2}-2}=\frac{y}{x-8}$, i.e., $y_{1}-y_{2}=\frac{y}{x-8}\left(x_{1}-x_{2}\right)$.
Since points $A, B$ are on the hyperbola, we have $x_{1}^{2}-y_{1}^{2}=2, x_{2}^{2}-y_{2}^{2}=2$, subtracting these two equations gives
$\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}\right)=\left(y_{1}-y_{2}\right)\left(y_{1}+y_{2}\right)$, i.e., $\left(x_{1}-x_{2}\right)(x-4)=\left(y_{1}-y_{2}\right) y$.
Substituting $y_{1}-y_{2}=\frac{y}{x-8}\left(x_{1}-x_{2}\right)$ into the above equation, simplifying yields $(x-6)^{2}-y^{2}=4$.
When $A B$ is perpendicular to the $x$-axis, $x_{1}=x_{2}=2$, we find $M(8,0)$, which also satisfies the above equation. Therefore, the trajectory equation of point $M$ is $(x-$ $6)^{2}-y^{2}=4$.
(2) Assume there exists a fixed point $C(m, 0)$ on the $x$-axis such that $\overrightarrow{C A} \cdot \overrightarrow{C B}$ is a constant. When $A B$ is not perpendicular to the $x$-axis, let the equation of line $A B$ be $y=k(x-2)(k \neq \pm 1)$. Substituting into $x^{2}-y^{2}=2$ gives $\left(1-k^{2}\right) x^{2}+4 k^{2} x-\left(4 k^{2}+2\right)=0$.
Then $x_{1}, x_{2}$ are the two real roots of the above equation, so $x_{1}+x_{2}=\frac{4 k^{2}}{k^{2}-1}, x_{1} x_{2}=\frac{4 k^{2}+2}{k^{2}-1}$,
thus $\overrightarrow{C A} \cdot \overrightarrow{C B}=\left(x_{1}-m\right)\left(x_{2}-m\right)+k^{2}\left(x_{1}-2\right)\left(x_{2}-2\right)=\left(k^{2}+1\right) x_{1} x_{2}-\left(2 k^{2}+m\right)\left(x_{1}+x_{2}\right)+4 k^{2}$ $+m^{2}$
$$
=\frac{\left(k^{2}+1\right)\left(4 k^{2}+2\right)}{k^{2}-1}-\frac{4 k^{2}\left(2 k^{2}+m\right)}{k^{2}-1}+4 k^{2}+m^{2}=\frac{2(1-2 m) k^{2}+2}{k^{2}-1}+m^{2}=2(1-2 m)+\frac{4-4 m}{k^{2}-1}+
$$
$m^{2}$.
Since $\overrightarrow{C A} \cdot \overrightarrow{C B}$ is a constant independent of $k$, we have $4-4 m=0$, i.e., $m=1$, at this time $\overrightarrow{C A} \cdot \overrightarrow{C B}=-1$.
When $A B$ is perpendicular to the $x$-axis, the coordinates of points $A, B$ can be set as $(2, \sqrt{2}),(2,-\sqrt{2})$,
at this time $\overrightarrow{C A} \cdot \overrightarrow{C B}=(1, \sqrt{2}) \cdot(1,-\sqrt{2})=-1$. Hence, there exists a fixed point $C(1,0)$ on the $x$-axis such that $\overrightarrow{C A} \cdot \overrightarrow{C B}$ is a constant.