$20 \cdot 13$ In the circle on the right, chords $AB$ and $CD$ intersect at $E$ and are perpendicular to each other. If the lengths of segments $AE$, $EB$, and $ED$ are 2, 6, and 3, respectively, then the length of the diameter of the circle is
(A) $4 \sqrt{5}$.
(B) $\sqrt{65}$.
(C) $2 \sqrt{17}$.
(D) $3 \sqrt{7}$.
(E) $6 \sqrt{2}$.
[Solution] As shown in the figure, given $A E=2, E B=6, E D=3$, and by the intersecting chords theorem, the length of segment $C E$ is 4. The center of the circle $O$ is the intersection of the perpendicular bisectors of chords $A B$ and $C D$. Point $O$ is 4 units to the right and $\frac{1}{2}$ unit up from point $A$, thus
$$
\begin{aligned}
O A^{2} & =A M^{2}+O M^{2}=4^{2}+\left(\frac{1}{2}\right)^{2} \\
& =\left(\frac{1}{2} \sqrt{65}\right)^{2},
\end{aligned}
$$
$\therefore$ the length of the diameter $2 O A=\sqrt{65}$.
Therefore, the answer is $(B)$.