11. (2002 Beijing Spring College Entrance Examination) As shown in Figure $8-14$, let points $A$ and $B$ be two moving points on the parabola $y^{2}=4 p x(p>0)$, excluding the origin. Given that $O A \perp O B, O M \perp A B$, find the equation of the trajectory of point $M$, and explain what curve it represents.
11. Let the coordinates of $A$, $B$, and $M$ be $\left(\frac{y_{1}^{2}}{4 p}, y_{1}\right)$, $\left(\frac{y_{2}^{2}}{4 p}, y_{2}\right)$, and $(x, y)$, respectively.

From $\overrightarrow{O A} \perp \overrightarrow{O B}$, we get $\overrightarrow{O A} \cdot \overrightarrow{O B}=0$, which means $\frac{y_{1}^{2}}{4 p} \cdot \frac{y_{2}^{2}}{4 p} + y_{1} y_{2} = 0$,
simplifying to $y_{1} y_{2} = -16 p^{2}$.
Also, from $\overrightarrow{O M} \perp \overrightarrow{A B}$, we have $\frac{y_{2}^{2} - y_{1}^{2}}{4 p} x + (y_{2} - y_{1}) y = 0$,
simplifying to $\frac{y_{1} + y_{2}}{4 p} x + y = 0$.
Since $A$, $B$, and $M$ are collinear, then $\overrightarrow{A M} \parallel \overrightarrow{A B}$,
thus $\left(x - \frac{y_{1}^{2}}{4 p}\right)(y_{2} - y_{1}) - \left(\frac{y_{2}^{2}}{4 p} - \frac{y_{1}^{2}}{4 p}\right)(y - y_{1}) = 0$,

simplifying to $x - \frac{y_{1} + y_{2}}{4 p} y + \frac{y_{1} y_{2}}{4 p} = 0$.
Substituting (1) and (2) into (3), and simplifying, we get $x^{2} + y^{2} - 4 p x = 0$.
Since $A$ and $B$ are points other than the origin, $x \neq 0$.
Therefore, the equation of the trajectory of point $M$ is $x^{2} + y^{2} - 4 p x = 0 (x \neq 0)$,

which represents a circle with center $(2 p, 0)$ and radius $2 p$, excluding the origin.