$27 \cdot 6$ If point $A(a, 0)$ is outside the circle $O: x^{2}+y^{2}=1$, points $P, Q \in$ circle $O$, point $M, P$ are symmetric about point $A$, and $O P \perp O Q$, with $a>0$, then the maximum length of segment $M Q$ must be
(A) $\sqrt{4 a^{2}+4 \sqrt{2} a+2}$.
(B) $\sqrt{4 a^{2}-4 a+2}$.
(C) $\sqrt{a^{2}+a+1}$.
(D) $\sqrt{a^{2}-a+2}$.
(High School Mathematics Competition in Taiyuan City, Shanxi Province, China, 1990)
［Solution］Let the coordinates of point $P$ be $(\cos \theta, \sin \theta)$, then the coordinates of point $M$, which is symmetric to $P$ with respect to point $A$, are
$$
(2 a-\cos \theta,-\sin \theta),
$$

Since $O P \perp O Q$, the coordinates of point $Q$ are
$$
\left(\cos \left(\theta+90^{\circ}\right), \sin \left(\theta+90^{\circ}\right)\right)
$$

which is $(-\sin \theta, \cos \theta)$,
thus
$$
\begin{aligned}
|M Q|= & \sqrt{(2 a-\cos \theta+\sin \theta)^{2}+(-\sin \theta-\cos \theta)^{2}} \\
= & \sqrt{4 a^{2}+4 a(\sin \theta-\cos \theta)+2}, \\
& \sin \theta-\cos \theta=\sqrt{2} \sin \left(\theta-45^{\circ}\right),
\end{aligned}
$$

and
when $\theta=135^{\circ}$, $\quad(\sin \theta-\cos \theta)_{\max }=\sqrt{2}$.
Therefore,
$$
|M Q|_{\max }=\sqrt{4 a^{2}+4 \sqrt{2} a+2} \text {. }
$$

Hence, the answer is (A).