smc

As in solution 1, in triangle $ABC$ we have $AB=1+4=5$ and $BC=4-1=3$, thus by the Pythagorean theorem or pythagorean triples in general, we have $AC=4$. Let $r$ be the radius. Let $s$ be the perpendicular intersecting point $S$ and line $BC$. $AC=s$ because $s,$ both perpendicular radii, and $AC$ form a rectangle. We just have to find $AC$ in terms of $r$ and solve for $r$ now. From the Pythagorean theorem and subtracting to get lengths, we get $AC=s=4=\sqrt{(r+1{)}^2-(1-r{)}^2}+\sqrt{(r+4{)}^2-(4-r{)}^2}$, which is simply $4=\sqrt{4r}+\sqrt{16r}⟹\sqrt{r}=\frac{2}{3}⟹r=\boxed{\frac{4}{9}}.$