Question

Let $S$ be the set of circles in the coordinate plane that are tangent to each of the three circles with equations $x^2 + y^2 = 4,$ $x^2 + y^2 = 64,$ and $(x - 5)^2 + y^2 = 3.$ What is the sum of the areas of all circles in $S?$

$48\pi$

$68\pi$

$96\pi$

$102\pi$

$136\pi$

Solution:

The first two circles are concentric with radii $2$ and $8.$ A circle tangent to both either has radius $3$ with center at distance $5$ from the origin, or radius $5$ with center at distance $3$ from the origin.

The third circle has center $(5, 0)$ and radius $\sqrt3.$ Imposing tangency with it, exactly four of the radius- $3$ circles and four of the radius- $5$ circles work (two tangency types, each giving a symmetric pair).

The sum of the areas is $4 \cdot \pi(3)^2 + 4 \cdot \pi(5)^2 = 36\pi + 100\pi = 136\pi.$

Thus, the correct answer is E.

2022 AMC 12B Problem 21

Solution:

Problem 21 in Other Years