AMC 10 · 2022 · #22

Grade 8 arithmetic

tangent-circlescoordinate-geometryarea-circlescaseworksystems-of-equations caseworkidentify-subproblemsconvert-to-algebra ↑ Prerequisites: tangent-circlescoordinate-geometry

📏 Long solution 💡 4 insights

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 $S$ ?

Pick an answer.

(A)

$48\pi$

(B)

$68\pi$

(C)

$96\pi$

(D)

$102\pi$

(E)

$136\pi$

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: Three given circles: $C_1: x^2+y^2=4$ (center $O$, radius $2$), $C_2: x^2+y^2=64$ (center $O$, radius $8$), and $C_3: (x-5)^2+y^2=3$ (center $(5,0)$, radius $\sqrt{3}$). Find every circle $S$ tangent to all three. Sum the areas of those circles $S$.

Givens: $C_1$ and $C_2$ are concentric at the origin with radii $2$ and $8$; $C_3$ is centered at $(5,0)$ with radius $\sqrt{3}$; $C_3$ sits inside the annulus between $C_1$ and $C_2$ (since distance $5$ to origin and radius $\sqrt{3}$ keep it between the two concentric circles); Answer choices: (A) $48\pi$, (B) $68\pi$, (C) $96\pi$, (D) $102\pi$, (E) $136\pi$

Unknowns: $\sum_{S} \pi r_S^2$ over all qualifying circles $S$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem, #13 Convert to Algebra, #2 Make a Systematic List

Tool #1 (Diagram) — sketch the two concentric circles with the small offset $C_3$ inside the annulus; the picture immediately shows $S$ has to fit between $C_1$ and $C_2$ in one of two simple ways. Tool #9 (Easier) — solve the easier sub-problem first: "which circles are tangent to BOTH concentric circles?" That forces just two radii ($3$ or $5$). Tool #7 (Subproblems) — split into two cases (radius $3$ or $5$) and within each, two more sign-choices for the $C_3$ tangency. Tool #2 (List) — systematically list each tangency configuration to be sure none are missed. Tool #13 (Algebra) — set up the distance equation for the center to confirm each case yields valid circles.

Execute — Answer: E

#1 Draw a Diagram 7.G.B.4 Step 1

Draw $C_1$ (radius $2$) and $C_2$ (radius $8$) sharing center $O$.
They form an annulus of width $8 - 2 = 6$.
Plot $C_3$ at $(5,0)$ with radius $\sqrt{3} \approx 1.73$.
Since $5 - \sqrt{3} \approx 3.27 > 2$ and $5 + \sqrt{3} \approx 6.73 < 8$, the whole circle $C_3$ lives inside the annulus.

$$\text{annulus width} = 8 - 2 = 6, \quad C_3 \subset \text{annulus}$$

💡 Grade 7 — recognize a circle from $(x-a)^2 + (y-b)^2 = r^2$ form and sketch it.

#9 Solve an Easier Related Problem 7.G.B.4 Step 2

Easier sub-problem: which circles $S$ (radius $r$, center $C$) are tangent to both concentric circles $C_1, C_2$?
Let $d = |OC|$.
There are exactly two ways: (i) $S$ sits in the annulus, externally tangent to $C_1$ and internally inside $C_2$, giving $d = r + 2$ and $d = 8 - r$, so $r = 3, d = 5$.
(ii) $S$ contains $C_1$ and sits inside $C_2$, giving $d = r - 2$ and $d = 8 - r$, so $r = 5, d = 3$.

$$r = 3, \;d = 5 \quad \text{or} \quad r = 5, \;d = 3$$

💡 Grade 7 — using tangency between two circles as a distance condition $d = r_1 \pm r_2$.

#7 Identify Subproblems 8.G.B.8 Step 3

So every solution circle has center on either $x^2 + y^2 = 25$ (radius $3$ case) or $x^2 + y^2 = 9$ (radius $5$ case).
Now add tangency to $C_3$: $\text{dist}(C, (5,0)) = r \pm \sqrt{3}$.
Squaring, $(x-5)^2 + y^2 = (r \pm \sqrt{3})^2$.

$$(x-5)^2 + y^2 = (r \pm \sqrt{3})^2$$

💡 Grade 8 — translate the third tangency into a distance equation in the coordinate plane.

#13 Convert to Algebra 8.G.B.8 Step 4

Case A ($r=3$): expand $(x-5)^2 + y^2 = x^2 - 10x + 25 + y^2 = 25 + (-10x) + 25 = 50 - 10x$ (using $x^2+y^2=25$).
So $50 - 10x = (3 \pm \sqrt{3})^2$.
Two distinct right-hand sides give two distinct $x$ values: $x = \tfrac{50 - (3+\sqrt{3})^2}{10}$ and $x = \tfrac{50 - (3-\sqrt{3})^2}{10}$.
For each $x$, the center constraint $y^2 = 25 - x^2$ gives $y = \pm\sqrt{25 - x^2}$ — two centers symmetric across the $x$-axis.
Total $2 \times 2 = 4$ circles with $r = 3$.

$$50 - 10x = (3 \pm \sqrt{3})^2 \;\Rightarrow\; 4 \text{ circles}$$

💡 Grade 8 — two sign choices times two reflections across the $x$-axis.

#13 Convert to Algebra 8.G.B.8 Step 5

Case B ($r=5$): same expansion, but now $x^2+y^2=9$.
$(x-5)^2 + y^2 = 9 - 10x + 25 = 34 - 10x = (5 \pm \sqrt{3})^2$.
Again two distinct $x$, and for each $y = \pm\sqrt{9 - x^2}$.
Quick sanity: $(5+\sqrt{3})^2 \approx 45.3$ gives $x \approx -1.13$, and $(5-\sqrt{3})^2 \approx 10.7$ gives $x \approx 2.33$.
Both $|x| < 3$, so $y$ is real.
$4$ more circles with $r = 5$.

$$34 - 10x = (5 \pm \sqrt{3})^2 \;\Rightarrow\; 4 \text{ circles}$$

💡 Grade 8 — same recipe applied to the second radius.

#2 Make a Systematic List 7.G.B.4 Step 6

Systematic list of areas.
$4$ circles with $r=3$ contribute $4 \cdot \pi \cdot 3^2 = 36\pi$.
$4$ circles with $r=5$ contribute $4 \cdot \pi \cdot 5^2 = 100\pi$.
Total $36\pi + 100\pi = 136\pi$, choice (E).

$$36\pi + 100\pi = 136\pi \;\Rightarrow\; \textbf{(E)}$$

💡 Grade 7 — area of a circle $= \pi r^2$, summed across the eight circles.

[1] #1 7.G.B.4 Draw $C_1$ (radius $2$) and $C_2$ (radius $8$) sharing center $O$. They form an

[2] #9 7.G.B.4 Easier sub-problem: which circles $S$ (radius $r$, center $C$) are tangent to bo

[3] #7 8.G.B.8 So every solution circle has center on either $x^2 + y^2 = 25$ (radius $3$ case)

[4] #13 8.G.B.8 Case A ($r=3$): expand $(x-5)^2 + y^2 = x^2 - 10x + 25 + y^2 = 25 + (-10x) + 25

[5] #13 8.G.B.8 Case B ($r=5$): same expansion, but now $x^2+y^2=9$. $(x-5)^2 + y^2 = 9 - 10x +

[6] #2 7.G.B.4 Systematic list of areas. $4$ circles with $r=3$ contribute $4 \cdot \pi \cdot 3

Review

Reasonableness: Sanity. The two concentric circles alone force radius $\in \{3, 5\}$ — this is a clean geometric pinch and easy to double-check by drawing. The third circle $C_3$ then permits two tangency sign-choices and the configuration is symmetric in $y$, multiplying by $2$. So we expect a count divisible by $4$ per radius — and we got exactly $4 + 4 = 8$ circles. Each $x$-value computed lies inside the corresponding locus circle ($|x|<5$ for case A, $|x|<3$ for case B), so all $y$'s are real and the eight circles truly exist. Sum $36\pi + 100\pi = 136\pi$ matches (E). Choices $48\pi, 68\pi, 96\pi, 102\pi$ are decoys for forgetting one case or one sign.

Alternative: Tool #3 (Eliminate). All five answer choices are multiples of $\pi$ between $48$ and $136$, and the area must be of the form $4k_1 \cdot 9 + 4 k_2 \cdot 25 = 36 k_1 + 100 k_2$ for integer $k_1, k_2 \in \{0,1\}$ counting cases. Only $(k_1, k_2) = (1, 1)$ gives $136\pi$, matching (E); the smaller answers would correspond to missing a case entirely.

CCSS standards used (min grade 8)

7.G.B.4 Know the formulas for the area and circumference of a circle and use them to solve problems (Identifying circles from their equations and computing total area $4\pi \cdot 9 + 4\pi \cdot 25 = 136\pi$.)
8.G.B.8 Apply the Pythagorean theorem to find distance between two points in a coordinate system (Using the distance equation $(x-5)^2 + y^2 = (r \pm \sqrt{3})^2$ to enforce tangency with $C_3$.)

⭐ This AMC 10 problem only needs Grade 8 coordinate-plane distance you already know — drawing the two same-center circles immediately pins the unknown radius to $3$ or $5$, and each of those two cases gives $4$ valid circles (two sign-choices for tangency with the third circle, times mirror symmetry). Eight circles in total, area $4\pi(9) + 4\pi(25) = 136\pi$.