Circles with centers $A$ and $B$ have radii $3$ and $8$, respectively. A common internal tangent intersects the circles at $C$ and $D$, respectively. Lines $AB$ and $CD$ intersect at $E$, and $AE=5$. What is $CD$? $\mathrm{(A)}\ 13\qquad$ $\mathrm{(B)}\ \frac{44}{3}\qquad$ $\mathrm{(C)}\ \sqrt{221}\qquad$ $\mathrm{(D)}\ \sqrt{255}\qquad$ $\mathrm{(E)}\ \frac{55}{3}$
Solution
Connect $AC$ and $BD$; both are perpendicular to the tangent $CD$, so $AC\parallel BD$. Since $C,D$ are on opposite sides of $E$, $\triangle ACE\sim\triangle BDE$ with ratio $3:8$.
$EB=\frac83AE=\frac{40}3$, so $AB=5+\frac{40}3=\frac{55}3$.
Through $A$ draw $AF\parallel CD$ meeting line $BD$ at $F$ (beyond $D$). Then $ACDF$ is a rectangle, so $AF=CD$ and $BF=BD+DF=8+3=11$.
In right triangle $AFB$, $CD=AF=\sqrt{AB^2-BF^2}=\sqrt{\frac{3025}9-121}=\sqrt{\frac{1936}9}=\frac{44}3$.
Circles with centers $A$ and $B$ have radii $4$ and $9$, respectively. A common internal tangent intersects the circles at $C$ and $D$, respectively. Lines $AB$ and $CD$ intersect at $E$, and $AE=5$. What is $CD$?
$\mathrm{(A)}\ 13\qquad$ $\mathrm{(B)}\ \frac{44}{3}\qquad$ $\mathrm{(C)}\ \sqrt{221}\qquad$ $\mathrm{(D)}\ \sqrt{255}\qquad$ $\mathrm{(E)}\ \frac{55}{3}$ 
