2022 AMC 10 B Problem 22

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To help you master 2022 AMC 10 B Problem 22, this page includes:

The original AMC problem to try, plus the correct answer

Clear, step-by-step solutions using different methods

Key ideas, useful formulas, and takeaways you can apply to similar problems

Related problems from past AMC exams, plus new practice problems that build the same skills

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2022 AMC10 B Problem 22

Solution

Solution 1

Locus of Centers Method

By exploiting the concentric circles, we can find the exact radius and distance to the origin for any tangent circle. This turns the third tangency condition into finding the intersections of two locus circles.

Two circles with radii $r_{1}$ and $r_{2}$ are tangent if the distance between their centers is exactly $r_{1} + r_{2}$ (external tangency) or $∣ r_{1} - r_{2} ∣$ (internal tangency).

Step 1: Identify the properties of the three given circles

Extract the center coordinates and radii from the given circle equations by recognizing the standard form $(x - h)^{2} + (y - k)^{2} = r^{2}$ .

C_{1} : Center O (0, 0), r_{1} = 2

C_{2} : Center O (0, 0), r_{2} = 8

C_{3} : Center Q (5, 0), r_{3} = 3

Step 2: Determine conditions for tangency to the concentric circles

Let the unknown circle have center $P$ and radius $r$ . Because $C_{1}$ and $C_{2}$ are centered at the origin, the distance $d$ from the origin to $P$ can be determined. For the new circle to be tangent to $C_{1}$ , $d$ must be the sum or the absolute difference of their radii. The same logic applies to $C_{2}$ .

d = r + 2 or d = ∣ r - 2∣

d = r + 8 or d = ∣ r - 8∣

Step 3: Solve for the possible values of radius and distance

We set the possible expressions for $d$ equal to each other to find valid combinations of $r$ and $d$ , keeping in mind that both quantities must be non-negative. Testing the sum against the absolute difference gives one valid solution, as does testing the two absolute differences against each other. Other combinations, such as equating the two sums, lead to contradictions.

r + 2 = 8 - r ⟹ 2 r = 6 ⟹ r = 3, d = 5

r - 2 = 8 - r ⟹ 2 r = 10 ⟹ r = 5, d = 3

Step 4: Establish the two families of tangent circles

The valid solutions show that any circle tangent to both $C_{1}$ and $C_{2}$ must belong to one of two families based on its radius and the location of its center.

Family 1: Radius r = 3, center P lies on a circle of radius 5 around O .

Family 2: Radius r = 5, center P lies on a circle of radius 3 around O .

Step 5: Determine tangency conditions for the third circle

For a circle from either family to be tangent to $C_{3}$ , the distance from its center $P$ to $Q$ must be either the sum or the absolute difference of their radii. Therefore, the valid centers $P$ are the intersection points of the locus circle from Step 4 and a target circle centered at $Q$ .

PQ = r + 3 or PQ = ∣ r - 3 ∣

Step 6: Count the valid circles in the first family

For Family 1, the target distance to $Q$ is calculated. This means $P$ must lie on one of two circles centered at $Q$ . We verify that the locus circle centered at the origin intersects both of these target circles by checking if the distance between their centers, which is 5, is strictly between the sum and difference of their radii. Since all strict inequalities hold, each target distance yields exactly 2 intersection points, giving 4 distinct valid circles. We then calculate their total area.

PQ \in {3 + 3, 3 - 3}

PQ = 3 + 3 ⟹ ∣5 - (3 + 3) ∣ < 5 < 5 + (3 + 3)

PQ = 3 - 3 ⟹ ∣5 - (3 - 3) ∣ < 5 < 5 + (3 - 3)

Area_{1} = 4 \times (π \cdot 3^{2}) = 36 π

Step 7: Count the valid circles in the second family

Similarly, for Family 2, we test the two target distances to $Q$ . The center distance between the origin and $Q$ remains 5. We test the circle intersection inequalities for both target radii. The inequalities hold strictly, so we again find 2 intersection points for each distance, yielding 4 more valid circles. We calculate their combined area.

PQ \in {5 + 3, 5 - 3}

PQ = 5 + 3 ⟹ ∣3 - (5 + 3) ∣ < 5 < 3 + (5 + 3)

PQ = 5 - 3 ⟹ ∣3 - (5 - 3) ∣ < 5 < 3 + (5 - 3)

Area_{2} = 4 \times (π \cdot 5^{2}) = 100 π

Step 8: Calculate the total area

Sum the areas of the valid circles found in both families to obtain the final answer.

Total Area = 36 π + 100 π = 136 π

Correct answer: E. $136 π$
The sum of the areas is 136\pi, which corresponds to answer choice E. We found exactly 8 tangent circles, which is the expected maximum number of solutions to the Problem of Apollonius (finding circles tangent to three given circles). The use of strict triangle inequalities ensures that no locus circles degenerate into points or merge into a single tangent point.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Distance Condition for Tangent Circles:

d = r_{1} + r_{2} (external) or d = ∣ r_{1} - r_{2} ∣ (internal)

$d$ is the distance between the centers of the two circles; $r_{1}$ and $r_{2}$ are their radii. This is used to translate the geometric requirement of tangency into algebraic equations involving the distances to the center of the unknown circle.

Intersection of Two Circles:

∣ R_{1} - R_{2} ∣ < D < R_{1} + R_{2}

$R_{1}$ and $R_{2}$ are the radii of the two circles, and $D$ is the distance between their centers. This strict triangle inequality proves that the locus of possible centers $P$ and the required distance to $Q$ yield exactly two distinct intersection points in every case, confirming the number of valid circles.

Key Takeaway: Remember This Strategy for Next Time

A problem involving circles tangent to concentric circles.

Try: Find the distance from the concentric center to the center of the new circle.

Why: Concentric circles share the same center, which reduces the 2D tangency problem into a 1D system of distance equations, creating a simple circular locus of possible centers.

You need to know how many points satisfy two distance conditions simultaneously.

Try: Treat the two distance conditions as two circles and check their intersection using the triangle inequality.

Why: A point at distance $R_{1}$ from $A$ and $R_{2}$ from $B$ exists if and only if the circles centered at $A$ and $B$ with those respective radii intersect.

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2022 AMC 10 B Problem 22

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2022 AMC10 B Problem 22