9. (16 points) In the Cartesian coordinate system $x O y$, $\angle M O N=\frac{\pi}{3}$, and side $M N$ moves on the line $l: x=3$. Find the equation of the locus of the circumcenter of $\triangle M O N$.

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9. Let $\angle x O N=\theta$, and point $N$ is above $M$. Then $M\left(3,3 \tan \left(\theta-\frac{\pi}{3}\right)\right), N(3,3 \tan \theta)$.
Let the circumcenter of $\triangle M O N$ be $P(x, y)$.
By the midpoint formula, the midpoint of $O N$ is
$C\left(\frac{3}{2}, \frac{3}{2} \tan \theta\right)$.
By the property of the circumcenter,
$y=\frac{3}{2}\left(\tan \theta+\tan \left(\theta-\frac{\pi}{3}\right)\right)$
$\Rightarrow \tan \theta+\tan \left(\theta-\frac{\pi}{3}\right)=\frac{2}{3} y$.
By $P C \perp O N$ we get
$\frac{y-\frac{3}{2} \tan \theta}{x-\frac{3}{2}} \tan \theta=-1$
$\Rightarrow \tan \left(\theta-\frac{\pi}{3}\right) \cdot \tan \theta=\frac{2}{3}\left(\frac{3}{2}-x\right)$.
Also, $\sqrt{3}=\tan \frac{\pi}{3}=\tan \left(\theta-\left(\theta-\frac{\pi}{3}\right)\right)$, then
$\tan \theta-\tan \left(\theta-\frac{\pi}{3}\right)$
$=\sqrt{3}\left(1+\tan \theta \cdot \tan \left(\theta-\frac{\pi}{3}\right)\right)$.
Squaring both sides of the above equation and substituting equations (1) and (2) we get $\frac{(x-4)^{2}}{4}-\frac{y^{2}}{12}=1$.