Example 3 Given $O$ is the origin, the side $AB$ of $\triangle ABO$ moves on the line $l: x=3$.
(1) Let the circumcenter of $\triangle ABO$ be $P\left(x_{P}, y_{P}\right), A(3, t_{1}), B\left(3, t_{2}\right)$, express the value of $t_{1} t_{2}$ in terms of $x_{P}$;
(2) If $\angle AOB=\frac{\pi}{3}$, find the equation of the locus of the circumcenter $P$ of $\triangle AOB$.
Solution: (1) The midpoint of $O A$ is $\left(\frac{3}{2}, \frac{t_{1}}{2}\right)$, the equation of the perpendicular bisector of line segment $O A$ is $y-\frac{t_{1}}{2}=-\frac{3}{t_{1}}\left(x-\frac{3}{2}\right)(t \neq 0)$,
i.e., $x+\frac{1}{3} t_{1} y-\frac{1}{6} t_{1}^{2}-\frac{3}{2}=0$.
The equation of the perpendicular bisector of line segment $A B$ is $y=\frac{t_{1}+t_{2}}{2}$. Solving (1) and (2) together yields $t_{1} t_{2}=9-6 x$,
If $t_{1}=0$, then $x_{P}=\frac{3}{2}$, i.e., $0=t_{1} t_{2}=9-6 x$,
$$
\therefore t_{1} t_{2}=9-6 x_{P} \text {. }
$$
(2) Let the circumcenter of $\triangle A B O$ be $P(x, y)$, given $\angle A O B=\frac{\pi}{3}$, by the Law of Sines we get
$$
\begin{array}{l}
|A B|=\left|t_{1}-t_{2}\right|=\sqrt{\left(t_{1}+t_{2}\right)^{2}-4 t_{1} t_{2}}= \\
\sqrt{(2 y)^{2}-4(9-6 x)}=\sqrt{3\left(x^{2}+y^{2}\right)}, \\
\therefore \frac{(x-4)^{2}}{4}-\frac{y^{2}}{12}=1(x \leqslant 2) .
\end{array}
$$