2012 AMC 10A Problem 11

1. The circles touch each other, so the distance between the centers is the sum of the radii: $AB = 5 + 3 = 8$.
2. A line tangent to both circles on the same side slopes toward the small circle and crosses the line of centers beyond it. So $C$ is past $B$, and $AC = AB + BC = 8 + BC$.
3. Let the line touch the big circle at $P$ and the small circle at $Q$. Draw the radii $AP$ and $BQ$. A radius drawn to the point where a tangent line touches meets that line at a right angle, so $\angle APC$ and $\angle BQC$ are both $90^\circ$.
4. Triangles $APC$ and $BQC$ each have a right angle and share the angle at $C$, so they have the same shape (they are similar). Matching sides are in the same ratio:
$$\frac{AC}{BC} = \frac{AP}{BQ} = \frac{5}{3}.$$
5. Put in $AC = BC + 8$ and cross-multiply:
$$3(BC + 8) = 5\,BC \quad\Longrightarrow\quad 3\,BC + 24 = 5\,BC \quad\Longrightarrow\quad 24 = 2\,BC \quad\Longrightarrow\quad BC = 12.$$
6. Check: $AC = 20$, and $20 : 12 = 5 : 3$. It matches.

The answer is $\boxed{\textbf{(D)}\ 12}$.

Whenever one line is tangent to two circles, draw the radii to the touching points. Both radii make right angles with the same line, so they are parallel, and you get two similar right triangles. Their size ratio is the ratio of the radii, and the distance between the centers gives the second equation you need.

Look at the difference instead. Since $AC : BC = 5 : 3$, write $AC = 5k$ and $BC = 3k$. Their difference is $AB = 8$, so $2k = 8$ and $k = 4$. Then $BC = 3 \times 4 = 12$.