(11) (20 points) In the plane rectangular coordinate system $x O y$, the side length of rhombus $A B C D$ is 4, and $|O B|=|O D|=6$.
(1) Prove that $|O A| \cdot|O C|$ is a constant;
(2) When point $A$ moves on the semicircle $M:(x-2)^{2}+$ $y^{2}=4(2 \leqslant x \leqslant 4)$, find the trajectory of point $C$.
11 (1) Since $|O B|=|O D|,|A B|=$ $|A D|=|C B|=|C D|$, points $O$, $A$, and $C$ are collinear.

As shown in the figure, connect $B D$, then $B D$ is the perpendicular bisector of segment $A C$, and let the foot of the perpendicular be $K$. Therefore, we have
$$
\begin{aligned}
& |O A| \cdot|O C| \\
= & (|O K|-|A K|)(|O K|+|A K|) \\
= & |O K|^{2}-|A K|^{2} \\
= & \left(|O B|^{2}-|B K|^{2}\right)-\left(|A B|^{2}-|B K|^{2}\right) \\
= & |O B|^{2}-|A B|^{2} \\
= & 6^{2}-4^{2}=20 \text{ (constant). }
\end{aligned}
$$
(2) Let $C(x, y)$ and $A(2+2 \cos \alpha, 2 \sin \alpha)$, where $\alpha=\angle X M A$ $\left(-\frac{\pi}{2} \leqslant \alpha \leqslant \frac{\pi}{2}\right)$, then $\angle X O C=\frac{\alpha}{2}$.

Since $|O A|^{2}=(2+2 \cos \alpha)^{2}+(2 \sin \alpha)^{2}=8(1+\cos \alpha)=16 \cos ^{2} \frac{\alpha}{2}$, we have $|O A|=4 \cos \frac{\alpha}{2}$.
From the conclusion in (1), we get $|O C| \cos \frac{\alpha}{2}=5$.
Thus, $x=|O C| \cos \frac{\alpha}{2}=5$.
Therefore, $y=|O C| \sin \frac{\alpha}{2}=5 \tan \frac{\alpha}{2} \in[-5,5]$.
Hence, the trajectory of point $C$ is a line segment, with the coordinates of its two endpoints being $(5,5)$ and $(5,-5)$.