2022 AMC 12B Problems/Problem 21

Problem

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 $S$ ?

$\textbf{(A)}~48\pi\qquad\textbf{(B)}~68\pi\qquad\textbf{(C)}~96\pi\qquad\textbf{(D)}~102\pi\qquad\textbf{(E)}~136\pi\qquad$

Solution

$[asy] import geometry; unitsize(0.5cm); void dc(pair x, pen p) { pair y = intersectionpoints(circle((0,0),8),(0,0)--1000*x)[0]; draw(circle(x, abs(x-y)),p); } pair O1 = (0,0),O2=(5,0),P1=intersectionpoints(circle(O1,5),circle(O2,3+sqrt(3)))[0],P2=intersectionpoints(circle(O1,3),circle(O2,5+sqrt(3)))[0],P3=intersectionpoints(circle(O1,5),circle(O2,3-sqrt(3)))[0],P4=intersectionpoints(circle(O1,3),circle(O2,5-sqrt(3)))[0]; draw(circle(O1,2)); draw(circle(O1,8)); draw(circle(O2,sqrt(3))); dc(P1,blue); dc(P2,red); dc(P3,darkgreen); dc(P4,brown); [/asy]$ The circles match up as follows: Case 1 is brown, Case 2 is blue, Case 3 is green, and Case 4 is red.[/center] Let $x^2 + y^2 = 64$ be circle $O$, $x^2 + y^2 = 4$ be circle $P$, and $(x-5)^2 + y^2 = 3$ be circle $Q$. All the circles in S are internally tangent to circle $O$. There are four cases with two circle belonging to each:

[*] $P$ and $Q$ are internally tangent to S. [*] $P$ and $Q$ are externally tangent to S. [*] $P$ is externally and Circle $Q$ is internally tangent to S. [*] $P$ is internally and Circle $Q$ is externally tangent to S.

Consider Cases 1 and 4 together. Since circles $O$ and $P$ have the same center, the line connecting the center of $S$ and the center of $O$ will pass through both the tangency point of $S$ and $O$ and the tangency point of $S$ and $P$. This line will be the diameter of $S$ and have length $r_P + r_O = 10$. Therefore the radius of $S$ in these cases is 5.

Consider Cases 2 and 3 together. Similarly to Case 1 and 4, the line connecting the center of $S$ to the center of $O$ will pass through the tangency points. This time however, the diameter of $S$ will have length $r_P-r_O=6$. Therefore, the radius of $S$ in these cases is 3.

The set of circles S consists of 8 circles - 4 of which have radius 5 and 4 of which have radius 3. The total area of all circles in S is $4(5^2\pi + 3^2\pi) = 136\pi \Rightarrow \boxed{\textbf{(E)}}$.

-naman12

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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All AMC 12 Problems and Solutions

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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All AMC 10 Problems and Solutions

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2022 AMC 12B Problems/Problem 21

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