AMC 10 · 2007 · #13
Grade 7 geometry-2dTwo circles of radius 2 are centered at (2,0) and at (0,2). What is the area of the intersection of the interiors of the two circles?
Pick an answer.
AMC 10 2007 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.
Toolkit + CCSS Solution
Understand
Restated: Two circles, each of radius $2$, are centered at $(2,0)$ and $(0,2)$. Find the area of the region that lies inside both circles at once.
Givens: Circle 1 has center $(2,0)$ and radius $2$; Circle 2 has center $(0,2)$ and radius $2$; The two radii are equal; Answer choices: (A) $\pi-2$, (B) $\frac{\pi}{2}$, (C) $\frac{\pi\sqrt3}{3}$, (D) $2(\pi-2)$, (E) $\pi$
Unknowns: The area of the overlap (the lens-shaped region interior to both circles)
Understand
Restated: Two circles, each of radius $2$, are centered at $(2,0)$ and $(0,2)$. Find the area of the region that lies inside both circles at once.
Givens: Circle 1 has center $(2,0)$ and radius $2$; Circle 2 has center $(0,2)$ and radius $2$; The two radii are equal; Answer choices: (A) $\pi-2$, (B) $\frac{\pi}{2}$, (C) $\frac{\pi\sqrt3}{3}$, (D) $2(\pi-2)$, (E) $\pi$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
The overlap is a curved lens, and the fastest way to see its shape is to draw both circles and mark where they cross (Tool #1). The lens has no simple area formula on its own, so break it into pieces we can measure: each half is a circular sector with a triangle removed (Tool #7 Identify Subproblems). Finally, since the problem is multiple choice, match the computed value against the five options (Tool #3).
Execute — Answer: D
6.NS.C.8 Step 1 Find where the circles cross
- Draw both circles.
- A point lies on both circles when it is exactly $2$ away from each center.
- Test $(0,0)$: center $(2,0)$ is a horizontal step of $2$ away and center $(0,2)$ is a vertical step of $2$ away, so $(0,0)$ sits on both circles.
- Test $(2,2)$: it is a vertical step of $2$ from $(2,0)$ and a horizontal step of $2$ from $(0,2)$, so it too sits on both.
- These two crossing points, $(0,0)$ and $(2,2)$, are the tips of the lens.
💡 The overlap is pinned between the two spots where the circle boundaries meet.
6.NS.C.8 Step 2 Split the lens into two equal segments
- The line through the two centers is an axis of symmetry, so the lens splits into two identical pieces.
- Focus on the half belonging to circle $1$, centered at $(2,0)$.
- Draw the two radii from $(2,0)$ to the tips $(0,0)$ and $(2,2)$.
- The radius to $(0,0)$ points straight left (horizontal), and the radius to $(2,2)$ points straight up (vertical), so the angle between them is a right angle, $90^\circ$.
- That half of the lens is a circular segment: the $90^\circ$ sector with the straight triangle cut off.
💡 Half the lens is a pie slice with the straight-edged triangle trimmed away, leaving just the curved sliver.
7.G.B.4 Step 3 Measure the sector and the triangle
- A $90^\circ$ sector is one quarter of a full circle.
- The full circle has area $\pi\cdot 2^2 = 4\pi$, so the sector is $\tfrac14\cdot 4\pi = \pi$.
- The triangle cut off has the two radii as its legs; both legs are $2$ and they meet at the right angle, so its area is $\tfrac12\cdot 2\cdot 2 = 2$.
- The curved sliver — one half of the lens — is the sector minus the triangle.
💡 A quarter circle minus its inscribed right triangle is exactly the leftover curved piece.
7.G.B.6 Step 4 Add the two halves
- One half of the lens has area $\pi - 2$.
- By the symmetry from the second step, the other half is identical, so the whole overlap is twice this.
- That gives $2(\pi - 2)$.
- Comparing with the choices, this matches (D) exactly, while (A) $\pi-2$ is only one half and the others do not fit.
- The area of the intersection is $\textbf{(D)}\ 2(\pi-2)$.
💡 The lens is two matching curved slivers, so double one sliver.
6.NS.C.8 Draw both circles. A point lies on both circles when it is exactly $2$ away from 6.NS.C.8 The line through the two centers is an axis of symmetry, so the lens splits into 7.G.B.4 A $90^\circ$ sector is one quarter of a full circle. The full circle has area $\ 7.G.B.6 One half of the lens has area $\pi - 2$. By the symmetry from the second step, t Review
Reasonableness: Estimate numerically: $2(\pi-2)\approx 2(3.1416-2)=2(1.1416)\approx 2.28$. This is a sliver, comfortably smaller than one whole circle's area $4\pi\approx 12.6$, which is what we expect for an overlap where the centers sit a good distance apart (the distance between centers is $2\sqrt2\approx 2.83$, more than the radius). A positive value a little above $2$ is sensible for a thin lens. Choice (A) $\pi-2\approx1.14$ is exactly half of our answer — the trap for stopping after one segment.
Alternative: Use the lens formula for two equal circles of radius $r$ whose centers are distance $d$ apart: $A = 2r^2\cos^{-1}\!\left(\tfrac{d}{2r}\right) - \tfrac{d}{2}\sqrt{4r^2-d^2}$. Here $r=2$ and $d=2\sqrt2$, so $\tfrac{d}{2r}=\tfrac{\sqrt2}{2}$ and $\cos^{-1}\!\left(\tfrac{\sqrt2}{2}\right)=\tfrac{\pi}{4}$. Then $A = 2\cdot4\cdot\tfrac{\pi}{4} - \sqrt2\cdot\sqrt{16-8} = 2\pi - \sqrt2\cdot 2\sqrt2 = 2\pi - 4 = 2(\pi-2)$, confirming (D).
CCSS standards used (min grade 7)
6.NS.C.8Solve problems by graphing points and finding distances between points sharing a coordinate (Locating the crossing points $(0,0)$ and $(2,2)$ and seeing the two radii are horizontal and vertical, hence perpendicular.)7.G.B.4Know and use the formulas for the area and circumference of a circle (Computing the $90^\circ$ sector as one quarter of the circle's area $\pi\cdot2^2=4\pi$.)6.G.A.1Find the area of triangles and other polygons (Computing the right triangle with legs $2$ and $2$ as area $\tfrac12\cdot2\cdot2=2$.)7.G.B.6Solve problems involving area of two-dimensional objects composed of simpler shapes (Getting each segment as sector minus triangle and doubling to get the full lens $2(\pi-2)$.)
⭐ To measure a curved lens where two circles overlap, cut it into pie-slice sectors with the straight triangles trimmed off, then add the leftover slivers.
⭐ To measure a curved lens where two circles overlap, cut it into pie-slice sectors with the straight triangles trimmed off, then add the leftover slivers.
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