2006 AMC 10A Problem 23

Draw the radii $AC$ and $BD$ to the points of tangency. A radius meets a tangent line at a right angle, so $\angle ACE = \angle BDE = 90^\circ$.

In right triangle $ACE$, the leg $AC = 3$ and the hypotenuse $AE = 5$, so the other leg is $CE = 4$ (a $3$-$4$-$5$ triangle).

Triangles $ACE$ and $BDE$ share vertical angles at $E$ and each has a right angle, so they are similar, with the ratio of corresponding sides equal to the ratio of radii, $BD : AC = 8 : 3$. Therefore
$$DE = \frac{8}{3} \cdot CE = \frac{8}{3} \cdot 4 = \frac{32}{3}.$$

Because $CD$ is an internal tangent, it crosses line $AB$ between the two circles, so $E$ lies between $C$ and $D$ and
$$CD = CE + DE = 4 + \frac{32}{3} = \frac{44}{3}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{44}{3}}$.

Any tangent problem starts by drawing the radius to the point of tangency, which manufactures a right angle. Two circles with a common tangent through a point on their line of centers then produce a pair of similar right triangles scaled by the radii. Internal versus external tangent only changes whether the two pieces of the tangent segment add or subtract.

• Scaling $AE = 5$ instead of $CE = 4$ and adding $5 + \tfrac{40}{3} = \tfrac{55}{3}$ (choice (E)), or treating the tangent as external and subtracting the two pieces.
• Using the wrong similarity ratio ($\tfrac{3}{8}$ instead of $\tfrac{8}{3}$), which makes $DE$ shorter than $CE$ even though the second circle is larger.