**The parabola $y^{2}=2 p x(p>0)$ has an inscribed triangle with one vertex at the origin, and the altitudes of the triangle all pass through the focus of the parabola. Find the equation of the circumcircle of this triangle.**
As shown in the figure, let $\triangle OAB$ be an inscribed triangle of a parabola. $AD$, $BE$, and $OC$ are three altitudes passing through the focus $F$. Let the coordinates of point $A$ be $\left(x_{1}, y_{1}\right)$, so $y_{1}=\sqrt{2 p x_{1}}$, and thus the coordinates of point $A$ are $\left(x_{1}, \sqrt{2 p x_{1}}\right)$. Since $AB$ is perpendicular to the $x$-axis and due to the symmetry of the parabola, the coordinates of point $B$ are $\left(x_{1}, -\sqrt{2 p x_{1}}\right)$. The slope of $OA$ is $k_{1} = \frac{\sqrt{2 p x_{1}}}{x_{1}}$, the slope of $BE$ is $k_{2} = \frac{-\sqrt{2 p x_{1}}}{x_{1} - \frac{p}{2}}$, and $k_{1} k_{2} = \frac{\sqrt{2 p x_{1}}}{x_{1}} \cdot \frac{-\sqrt{2 p x_{1}}}{x_{1} - \frac{p}{2}} = -1$. Therefore, $2 p = x_{1} - \frac{p}{2}$. So, $x_{1} = \frac{5 p}{2}$, $y_{1} = \sqrt{5} p$, and the coordinates of $A$ are $\left(\frac{5 p}{2}, \sqrt{5} p\right)$, and the coordinates of $B$ are $\left(\frac{5 p}{2}, -\sqrt{5} p\right)$. 

Let the equation of the circle passing through points $O$, $A$, and $B$ be $x^{2} + y^{2} + Dx + Ey + F = 0$. Since $O(0,0)$ is on the circle, $F = 0$. Since $A\left(\frac{5 p}{2}, \sqrt{5} p\right)$ is on the circle, we have $\frac{25}{4} p^{2} + 5 p^{2} + \frac{5 p}{2} D + \sqrt{5} p E = 0$. Since $B\left(\frac{5 p}{2}, -\sqrt{5} p\right)$ is on the circle, we have $\frac{25}{4} p^{2} + 5 p^{2} + \frac{5 p}{2} D - \sqrt{5} p E = 0$. Solving these equations, we get $D = -\frac{9 p}{2}$, $E = 0$, and $F = 0$. Therefore, the equation of the circumcircle of $\triangle OAB$ is: $x^{2} + y^{2} - \frac{9}{2} p x = 0$.

Alternatively, after obtaining the coordinates of $A$ and $B$ from the first part of the solution, the equation of the circumcircle can also be found as follows: The equation of the perpendicular bisector of $OA$ is:
$$
\sqrt{x^{2} + y^{2}} = \sqrt{\left(x - \frac{5}{2} p\right)^{2} + (y - \sqrt{5} p)^{2}},
$$
which simplifies to $5 x + 2 \sqrt{5} y = \frac{45}{4} p$;
The perpendicular bisector of $AB$ is $OC$, whose equation is $y = 0$. Solving the system of equations $\left\{\begin{array}{l}5 x + 2 \sqrt{5} y = \frac{45}{4} p, \\ y = 0,\end{array}\right.$, we get the coordinates of the center of the circle as $\left(\frac{9}{4} p, 0\right)$, and the radius $R = \frac{9}{4} p$. Therefore, the equation of the circle is $\left(x - \frac{9}{4} p\right)^{2} + y^{2} = \left(\frac{9}{4} p\right)^{2}$. Simplifying, we get $x^{2} + y^{2} - \frac{9}{2} p x = 0$.