jmc

Let $D$ and $F$ denote the centers of the circles. Let $C$ and $B$ be the points where the $x$-axis and $y$-axis intersect the tangent line, respectively. Let $E$ and $G$ denote the points of tangency as shown. We know that $AD=DE=2$, $DF=3$, and $FG=1$. Let $FC=u$ and $AB=y$. Triangles $FGC$ and $DEC$ are similar, so $$\frac{u}{1}=\frac{u+3}{2},$$ which yields $u=3$. Hence, $GC=\sqrt{8}$. Also, triangles $BAC$ and $FGC$ are similar, which yields $$\frac{y}{1}=\frac{BA}{FG}=\frac{AC}{GC}=\frac{8}{\sqrt{8}}=\sqrt{8}=\boxed{2\sqrt{2}}.$$