A circle centered at $A$ with a radius of 1 and a circle centered at $B$ with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. The radius of the third circle is $\text{(A) }\frac {1}{3} \qquad$ $ \text{(B) }\frac {2}{5} \qquad$ $ \text{(C) }\frac {5}{12} \qquad$ $ \text{(D) }\frac {4}{9} \qquad$ $ \text{(E) }\frac {1}{2}$
Solution
Let $X, Y,$ and $Z$ be the tangent points of the circles and the line.
Let $r$ be the radius of the small circle in the middle, by \underline{theorem 47}, we get $XY=2\sqrt{r}, YZ=2\sqrt{4r}, XZ=2\sqrt{4}$.
$2\sqrt{r}+2\sqrt{4r}=4$.
$r=\frac49$. The answer is D. In general, if the radii of those three circles are $a, b, c$, with $c$ is the smallest, then $$\frac1{\sqrt{c}}=\frac1{\sqrt{a}}+\frac1{\sqrt{b}},$$ this is summarized in theorem 48.
A circle centered at $A$ with radius $4$ and a circle centered at $B$ with radius $9$ are externally tangent. A third circle is tangent to the first two and to one of their common external tangents. What is the radius of the third circle?
$\text{(A) }\frac {1}{3} \qquad$ $ \text{(B) }\frac {2}{5} \qquad$ $ \text{(C) }\frac {5}{12} \qquad$ $ \text{(D) }\frac {4}{9} \qquad$ $ \text{(E) }\frac {1}{2}$ 
