11. As shown in the figure, in the Cartesian 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 $(x-2)^{2}+y^{2}=4(2 \leqslant x \leqslant 4)$, find the trajectory of point $C$.
(1) Let $B D$ intersect $O C$ at point $K$, then $|O A| \cdot|O C|=(|O K|-|A K|)(|O K|+|A K|)$ $=|O K|^{2}-|A K|^{2}=\left(|O B|^{2}-|B K|^{2}\right)-\left(|A B|^{2}-|B K|^{2}\right)$ $=|O B|^{2}-|A B|^{2}=36-16=20$. Therefore, $|O A| \cdot|O C|$ is a constant.
(2) Let $M$ be the center of the semicircle, $\angle A M X=\theta, \theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Thus, $A(2+2 \cos \theta, 2 \sin \theta)$, and by the collinearity of $O, A, C$, we have $C(2 k(1+\cos \theta), 2 k \sin \theta)$ $(k>0)$, substituting into $|O A| \cdot|O C|=4 k\left[(1+\cos \theta)^{2}+\sin ^{2} \theta\right]=8 k(1+\cos \theta)=20$ $\Rightarrow 2 k=\frac{5}{1+\cos \theta} \Rightarrow C\left(5, \frac{5 \sin \theta}{1+\cos \theta}\right)$. Since $\frac{\sin \theta}{1+\cos \theta}=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}=\tan \frac{\theta}{2} \in[-1,1]$,
the trajectory equation of point $C$ is $x=5(-5 \leqslant y \leqslant 5)$, which is a line segment connecting points $(5,5),(5,-5)$.