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 $S?$

$48\pi$

$68\pi$

$96\pi$

$102\pi$

$136\pi$

Answer: E

Concepts:tangent circles circle area casework

Difficulty rating: 2170

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 $\sqrt3.$ For each candidate radius $s\in\{3,5\},$ tangency requires the center's distance from $(5,0)$ to be $s+\sqrt3$ or $s-\sqrt3.$ Each of these two distance circles intersects the appropriate center-locus in two symmetric points. Hence exactly four radius- $3$ circles and four radius- $5$ circles work.

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

Answer: E

Difficulty rating: 2170

Solution:

Problem 21 in Other Years