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.

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