2022 AMC 12B Problem 21

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?

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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

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.

Tagged: tangent circles · circle area · casework