9. (16 points) Let point $F(1,0)$, line $l: x=-1$, line $l^{\prime} \perp l$ at point $P$, the perpendicular bisector of segment $P F$ intersects $l^{\prime}$ at point $Q$.
(1) Find the equation of the locus $C$ of point $Q$;
(2) Given point $H(1,2)$, a line passing through point $F$ and not perpendicular to the $x$-axis intersects curve $C$ at points $A$ and $B$, lines $A H$ and $B H$ intersect $l$ at points $M$ and $N$, respectively. Prove that the circle with diameter $M N$ must pass through a fixed point.
(2) According to the problem, let $l_{A B}: x=m y+1(m \neq 0)$.
Substitute into $y^{2}=4 x$, we get $y^{2}-4 m y-4=0$.
Let $A\left(\frac{y_{1}^{2}}{4}, y_{1}\right), B\left(\frac{y_{2}^{2}}{4}, y_{2}\right)$. Then
$y_{1}+y_{2}=4 m, y_{1} y_{2}=-4$.
Also, $H(1,2)$, thus,
$l_{A H}: y-2=\frac{4}{y_{1}+2}(x-1)$,
$l_{B H}: y-2=\frac{4}{y_{2}+2}(x-1)$.
Substitute $x=-1$ respectively, we get
$y_{M}=\frac{2\left(y_{1}-2\right)}{y_{1}+2}, y_{N}=\frac{2\left(y_{2}-2\right)}{y_{2}+2}$.
Then $y_{M} y_{N}=\frac{4\left(y_{1} y_{2}-2\left(y_{1}+y_{2}\right)+4\right)}{y_{1} y_{2}+2\left(y_{1}+y_{2}\right)+4}=-4$,
$y_{M}+y_{N}=4-8\left(\frac{1}{y_{1}+2}+\frac{1}{y_{2}+2}\right)$
$=4-\frac{8\left(\left(y_{1}+y_{2}\right)+4\right)}{y_{1} y_{2}+2\left(y_{1}+y_{2}\right)+4}$
$=-\frac{4}{m}$.
The equation of the circle with $M N$ as its diameter is
$$
\begin{array}{l}
(x+1)^{2}+\left(y-y_{M}\right)\left(y-y_{N}\right)=0 \\
\Rightarrow y^{2}-\left(y_{M}+y_{N}\right) y+y_{M} y_{N}+(x+1)^{2}=0 \\
\Rightarrow x^{2}+2 x-3+y^{2}+\frac{4}{m} y=0 .
\end{array}
$$

Let $y=0$, we solve to get $x=-3$ or 1.
Therefore, the circle with $M N$ as its diameter always passes through the fixed points $(-3,0)$ and $(1,0)$.