2011 AMC 12A Problem 11

Place $A = (-1, 0),$ $B = (1, 0),$ so their tangency point is the origin, the midpoint of $\overline{AB}.$ Then $C = (0, 1),$ since $C$ passes through the origin. The distance from $C$ to $A$ (and to $B$) is $\sqrt2.$ Two unit circles whose centers are $\sqrt2$ apart overlap in a lens of area $$\begin{aligned} 2\cos^{-1}\!\left(\tfrac{\sqrt2}{2}\right) \\ {}- \tfrac{\sqrt2}{2}\sqrt{4 - 2} \\ = 2 \cdot \tfrac{\pi}{4} - 1 = \tfrac{\pi}{2} - 1. \end{aligned}$$ Circles $A$ and $B$ meet only at the origin, so the two lenses do not overlap. The wanted area is $$\pi - 2\left(\tfrac{\pi}{2} - 1\right) = 2.$$ Thus, the correct answer is C.