Find the radius of the smallest circle

Let the three circles touch the straight line at $A, B, C$ respectively. Let the radius of the smallest circle be $r$ . Then, $|\overline {AB}|=\sqrt {48r}$ and $|\overline {BC}|=\sqrt {108r}$ . Therefore $|\overline {AC}|=\sqrt {r}(\sqrt {48}+\sqrt {108})=10\sqrt {3r}$ . Also, $|\overline {AC}|=\sqrt {(27+12)^2-(27-12)^2}=36$ . Therefore $10\sqrt {3r}=36$ or $r=\dfrac{36^2}{300}=\boxed {4.32}$