2007 AMC 10B Problem 13

P11-15~4 minPrint

Circles Coordinate Geometry Cut the figure into known shapes (triangles, rectangles, sectors)Exploit symmetry to reduce work or pair up objects

Two 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?

Show solution without answering

Key insight

Both circles pass through (0,0) and (2,2). The overlap is split by the line between those points into two equal pieces, each a quarter circle with a right triangle removed.

Solution

Sketch it first. One circle is centered at $(2,0)$ , the other at $(0,2)$ , both with radius $2$ .

1. Where do they cross? The point $(0,0)$ is $2$ away from $(2,0)$ and $2$ away from $(0,2)$ , so it is on both circles. So is $(2,2)$ : it is $2$ above $(2,0)$ and $2$ to the right of $(0,2)$ . The overlap is the football-shaped region between these two points.
2. Split the overlap. Draw the straight segment from $(0,0)$ to $(2,2)$ . It cuts the football into two pieces. Flipping the picture across the line $y = x$ swaps the two circles, so the two pieces are the same size. Find one and double it.
3. One piece. Take the piece that belongs to the circle centered at $C = (2,0)$ . From $C$ , the point $(0,0)$ is straight left and $(2,2)$ is straight up, so the two radii to them make a right angle.
- The quarter of that circle between those radii has area $\frac14 \times \pi \times 2^2 = \pi$ .
- The right triangle with corners $C$ , $(0,0)$ , $(2,2)$ has legs $2$ and $2$ , so area $\frac12 \times 2 \times 2 = 2$ .
- The piece is the quarter circle with the triangle removed: $\pi - 2$ .
4. Double it. The overlap is $2(\pi - 2)$ .

The answer is $\boxed{\textbf{(D)}\ 2(\pi -2)}$ .

Why this works

The overlap of two circles is always two "caps", one from each circle, glued along the line joining the crossing points. Each cap is a slice of the circle minus a triangle. Find the crossing points first, then the angle at each center; here the numbers were chosen so that angle is a neat $90^\circ$ .

Alternative approach

Use the square with corners $(0,0)$ , $(2,0)$ , $(2,2)$ , $(0,2)$ , which has area $4$ .

The quarter circle centered at $(2,0)$ that lies in this square has area $\pi$ . So does the quarter circle centered at $(0,2)$ .
Together these two quarter circles cover the whole square, and they overlap exactly in the football.
Adding the two quarter circles counts the square once plus the football one extra time: $\pi + \pi = 4 + \text{football}$ .
So the football is $2\pi - 4 = 2(\pi - 2)$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Each center is at distance $2$ from the origin and from $(2,2)$ , so both circles pass through $O = (0,0)$ and $P = (2,2)$ . These are the two intersection points, and the common region is the lens bounded by the two arcs from $O$ to $P$ .

The line $y = x$ through $O$ and $P$ is an axis of symmetry swapping the two circles, so the lens is two congruent circular segments, one from each circle, cut off by the chord $OP$ .

Take the circle centered at $C = (2,0)$ . The radii $CO$ and $CP$ point in the directions $(-2,0)$ and $(0,2)$ , which are perpendicular, so the chord $OP$ subtends a $90^\circ$ angle at $C$ . The segment area is therefore
$\text{quarter disk} - \text{right triangle} = \frac{1}{4}\pi\cdot 2^2 - \frac{1}{2}\cdot 2 \cdot 2 = \pi - 2.$

Doubling for the two segments gives the lens area $2(\pi - 2)$ .

The answer is $\boxed{\textbf{(D)}\ 2(\pi -2)}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 11: Circles
Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2, §4.3)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)