In the adjoining figure, $AB$ is tangent at $A$ to the circle with center $O$; point $D$ is interior to the circle; 
and $DB$ intersects the circle at $C$. If $BC=DC=3$, $OD=2$, and $AB=6$, then the radius of the circle is
$\textbf{(A) }3+\sqrt{3}\qquad \textbf{(B) }15/\pi\qquad \textbf{(C) }9/2\qquad \textbf{(D) }2\sqrt{6}\qquad  \textbf{(E) }\sqrt{22}$
Extend $\overline{BD}$ until it touches the opposite side of the circle, say at point $E.$ By power of a point, we have $AB^2=(BC)(BE),$ so $BE=\frac{6^2}{3}=12.$ Therefore, $DE=12-3-3=6.$
Now extend $\overline{OD}$ in both directions so that it intersects the circle in two points (in other words, draw the diameter of the circle that contains $OD$). Let the two endpoints of this diameter be $P$ and $Q,$ where $Q$ is closer to $C.$ Again use power of a point. We have $(PD)(DQ)=(CD)(DE)=3 \cdot 6=18.$ But if the radius of the circle is $r,$ we see that $PD=r+2$ and $DQ=r-2,$ so we have the equation $(r+2)(r-2)=18.$ Solving gives $r=\sqrt{22}.$