jmc

Let $C$ be the intersection of the horizontal line through $A$ and the vertical line through $B.$ In right triangle $ABC,$ we have $BC=3$ and $AB=5,$ so $AC=4.$ Let $x$ be the radius of the third circle, and $D$ be the center. Let $E$ and $F$ be the points of intersection of the horizontal line through $D$ with the vertical lines through $B$ and $A,$ respectively, as shown. In $△BED$ we have $BD=4+x$ and $BE=4-x,$ so $$D{E}^2=(4+x{)}^2-(4-x{)}^2=16x,$$and $DE=4\sqrt{x}.$ In $△ADF$ we have $AD=1+x$ and $AF=1-x,$ so $$F{D}^2=(1+x{)}^2-(1-x{)}^2=4x,$$and $FD=2\sqrt{x}.$ Hence, $$4=AC=FD+DE=2\sqrt{x}+4\sqrt{x}=6\sqrt{x}$$and $\sqrt{x}=\frac{2}{3},$ which implies $x=\boxed{\frac{4}{9}}.$