Question

Let $S$ be the set of circles in the coordinate plane that are tangent to each of the three circles with equations $\begin{gathered}x^{2}+y^{2}=4,\qquad x^{2}+y^{2}=64,\\ \text{and}\qquad (x-5)^{2}+y^{2}=3.\end{gathered}$ 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: 2390

Solution:

Call the concentric circles of radii $2$ and $8$ the inner and outer circles. Let a desired circle have radius $r$ and let its center be distance $d$ from the origin. It must be internally tangent to the outer circle, so $d+r=8.$

If it is externally tangent to the inner circle, then $d-r=2,$ giving $(r,d)=(3,5).$ If it contains the inner circle, then $r-d=2,$ giving $(r,d)=(5,3).$ Thus every desired circle has radius $3$ or $5.$

The third given circle has radius $\sqrt3$ and center $(5,0).$ For either value of $r,$ a desired circle may be internally or externally tangent to it, so the distance between their centers is $r-\sqrt3$ or $r+\sqrt3.$ In all four cases, if this distance is $q,$ then $|d-q|<5<d+q,$ so the circle of possible centers intersects the circle of radius $d$ about the origin in two points, symmetric across the $x$ -axis, as shown. Hence there are $4$ desired circles of each radius.

The total area is therefore $4(5^2\pi+3^2\pi)=136\pi.$ Thus, E is the correct answer.

2022 AMC 10B Problem 22

Answer: E

Difficulty rating: 2390

Solution:

Problem 22 in Other Years