2012 AMC 10A Problem 11

Since the circles are externally tangent, $AB = 5 + 3 = 8$. The common external tangent line meets the line of centers on the far side of the smaller circle, so $C$ lies beyond $B$ and $AC = AB + BC = 8 + BC$.

Let the tangent line touch the big circle at $P$ and the small circle at $Q$. Draw the radii $AP$ and $BQ$; both are perpendicular to the tangent line, so $AP \parallel BQ$. Triangles $APC$ and $BQC$ share the angle at $C$ and each has a right angle, so they are similar, with
$$\frac{AC}{BC} = \frac{AP}{BQ} = \frac{5}{3}.$$

Substituting $AC = BC + 8$:
$$\frac{BC + 8}{BC} = \frac{5}{3} \quad\Longrightarrow\quad 3BC + 24 = 5BC \quad\Longrightarrow\quad BC = 12.$$

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

Whenever a line is tangent to two circles, drawing the radii to the tangent points creates a pair of similar right triangles, because those radii are both perpendicular to the same line. The similarity ratio is the ratio of radii, and the center distance supplies the second equation. This "radii to tangent points" construction is the standard first move.

A homothety centered at $C$ sends the small circle to the big one with factor $\frac53$, so $CA = \frac53 CB$; with $CA - CB = 8$ this gives $\frac23 CB = 8$, $CB = 12$.

• Using AC = BC + 5 or + 3 instead of the full center distance 5 + 3 = 8, or placing C between the centers.
• Setting $\frac{AC}{BC} = \frac{3}{5}$ (ratio upside down), which gives $BC = -20$ or nonsense.