1994 AMC 12 Problem 17

The rectangle removes two congruent caps from the circle

The cap chord is at distance $\sqrt2$ from the center, so its half central angle is $45^\circ$

The rectangle is wider than the circle, so the common region is the circle with the top and bottom caps beyond distance $\sqrt2$ from the center removed. One cap has area $$\begin{aligned} 2^2\left(\frac{\pi}{4}\right) &-\sqrt2\sqrt{2^2-(\sqrt2)^2}\\ &=\pi-2. \end{aligned}$$ Therefore the common area is $4\pi-2(\pi-2)=2\pi+4.$ Thus the correct answer is D.