2007 AMC 10B Problem 13

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

Overlap of two circles is always two circular segments glued along the common chord, and a segment is "sector minus triangle." The coordinates here are chosen so the central angle is a clean $90^\circ$; in general, find the intersection points first and then read off the angle at each center.

The square with vertices $(0,0)$, $(2,0)$, $(2,2)$, $(0,2)$ lies inside both circles' quarter disks. The lens equals the two quarter disks minus the square counted twice, i.e. $2\pi - 2\cdot 4 + 4 = 2\pi - 4$; equivalently, the region of the square outside the lens consists of two pieces each of area $4 - \pi$, so the lens is $4 - 2(4 - \pi) = 2\pi - 4$.

• Computing only one circular segment (pi - 2) and forgetting that the lens consists of two of them.
• Using the distance between centers, $2\sqrt{2}$, as if it were the chord length, or assuming a $60^\circ$ central angle and producing the $\sqrt{3}$ distractor.