jmc

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

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