BEx ** Given a fixed point $A(0,2)$ and a circle $O: x^{2}+y^{2}=4$, a line $M A$ is tangent to circle $O$ at $A$, and $M$ is a moving point on the tangent line. $M Q$ is tangent to circle $O$ at point $Q$, as shown in the figure. Find the locus of the orthocenter $H$ of $\triangle M A Q$
Equation.
Analysis From the properties of the orthocenter, we know: $A H \perp M Q, Q H \perp$ $A M$. Also, from the relationship between the tangent and the circle, we know: $O Q \perp M Q, O A \perp$ $A M$. Therefore, $O A / / Q H, O Q / / A H$. This means quadrilateral $A O Q H$ is a parallelogram. Since $|O A|=|O Q|$, then $\square A O Q H$ is a rhombus.

Let $H(x, y)$, and $(x \neq 0)$, so $|A H|^{2}=$ $|O A|^{2}$, which means $x^{2}+(y-2)^{2}=4(x \neq 0)$. Therefore, the required trajectory equation is $x^{2}+(y-2)^{2}=4 .(x \neq 0)$