A circle of radius $r$ is surrounded by three circles, whose radii are 1, 2, and 3, all externally tangent to the inner circle and externally tangent to each other, as shown in the diagram below. What is $r$? $\textbf{(A) }\frac{1}{4}\qquad$ $\textbf{(B) }\frac{6}{23}\qquad$ $\textbf{(C) }\frac{3}{11}\qquad$ $\textbf{(D) }\frac{5}{17}\qquad$ $\textbf{(E) }\frac{3}{10}$
Solution
Let $O_1,O_2,O_3$ be the centers of the circles of radii $1,2,3$. Then $O_1O_2=3$, $O_1O_3=4$, $O_2O_3=5$, so $\angle O_2O_1O_3=90^\circ$. Place $O_1=(0,0)$, $O_2=(3,0)$, $O_3=(0,4)$ and let $P=(x,y)$ be the center of the small circle.
$PO_1=1+r$, $PO_2=2+r$, $PO_3=3+r$. Subtracting $x^2+y^2=(1+r)^2$ from $(x-3)^2+y^2=(2+r)^2$ gives $-6x+9=3+2r$, so $x=1-\frac r3$.
Subtracting $x^2+y^2=(1+r)^2$ from $x^2+(y-4)^2=(3+r)^2$ gives $-8y+16=8+4r$, so $y=1-\frac r2$.
Substituting into $x^2+y^2=(1+r)^2$: $\left(1-\frac r3\right)^2+\left(1-\frac r2\right)^2=(1+r)^2$, which simplifies to $23r^2+132r-36=0$, i.e. $(23r-6)(r+6)=0$. So $r=\frac6{23}$.
Three circles with radii $2$, $3$, and $10$ are pairwise externally tangent. A fourth circle, lying in the gap among them, is externally tangent to all three. What is the radius of the fourth circle?
What is $r$? $\textbf{(A) }\frac{1}{4}\qquad$ $\textbf{(B) }\frac{6}{23}\qquad$ $\textbf{(C) }\frac{3}{11}\qquad$ $\textbf{(D) }\frac{5}{17}\qquad$ $\textbf{(E) }\frac{3}{10}$ 
