The points $(-3,2)$ and $(-2,3)$ lie on a circle whose center is on the $x$-axis. What is the radius of the circle?
Let the center of the circle be $(x,0)$. Then we know the distance from the center to $(-3,2)$ and from the center to $(-2,3)$ are the same. Using the distance formula, we have

\begin{align*}
\sqrt{(x+3)^2+(0-2)^2}&=\sqrt{(x+2)^2+(0-3)^2}\\
\Rightarrow\qquad \sqrt{x^2+6x+9+4}&=\sqrt{x^2+4x+4+9}\\
\Rightarrow\qquad 6x&=4x\\
\Rightarrow\qquad x&=0\\
\end{align*}Now we know the center of the circle is $(0,0)$, and we need to find the radius. Use the distance formula once more: $$\sqrt{(0+3)^2+(0-2)^2}=\sqrt{3^2+(-2)^2}=\sqrt{9+4}=\boxed{\sqrt{13}}.$$