2006 AMC 10A Problem 23

1. Right angles. Draw the radii $AC = 3$ and $BD = 8$. A radius always meets a tangent line at a right angle at the touching point, so the angles at $C$ and at $D$ are right angles.
2. Find $CE$. Triangle $ACE$ has a right angle at $C$, leg $AC = 3$ and longest side $AE = 5$. This is a $3$-$4$-$5$ right triangle, so $CE = 4$.
3. Same shape. Triangles $ACE$ and $BDE$ each have a right angle, and their angles at $E$ are equal because they are vertical angles (the lines $AB$ and $CD$ cross at $E$). So the triangles have the same shape. The radius $8$ matches the radius $3$, so every side of $BDE$ is $\frac83$ times the matching side of $ACE$.
4. Find $DE$. $DE$ matches $CE$, so
$$DE = \frac83 \times 4 = \frac{32}{3}.$$
5. Add. The tangent crosses between the circles, so $E$ is between $C$ and $D$:
$$CD = CE + DE = 4 + \frac{32}{3} = \frac{44}{3}.$$

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

In a tangent problem, the first move is almost always to draw the radius to the touching point, which gives a right angle. Two circles with a shared tangent then give two right triangles of the same shape, scaled by the ratio of the radii. For an internal tangent (one that crosses between the circles), the two pieces of the tangent add.