$20 \cdot 25$ As shown, $AB$ is tangent to $\odot O$ at $A$, and point $D$ is inside the circle, with $DB$ intersecting the circle at $C$. If $BC=DC=3, OD=2, AB=6$, then the radius of the circle is
(A) $3+\sqrt{3}$.
(B) $15 / \pi$.
(C) $9 / 2$.
(D) $2 \sqrt{6}$.
(E) $\sqrt{22}$.
(27th American High School Mathematics Examination, 1976)
[Solution] Let $E$ be the intersection of the circle and the extension of $BD$, $FG$ be the diameter through $D$, and $r$ be the radius of the circle. Then
$BC \cdot BE = AB^2$, which means $3 \cdot (DE + 6) = 36$,
giving $DE = 6$.
Similarly, from $DE \cdot DC = DF \cdot DG$,
we have $18 = r^2 - 4$, yielding $r = \sqrt{22}$.
Therefore, the answer is $(E)$.