What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$ $\textbf{(A)}\ \pi+\sqrt{2}\qquad$ $\textbf{(B)}\ \pi+2\qquad$ $\textbf{(C)}\ \pi+2\sqrt{2}\qquad$ $\textbf{(D)}\ 2\pi+\sqrt{2}\qquad$ $\textbf{(E)}\ 2\pi+2\sqrt{2}$
Solution
In the first quadrant the equation is $\left(x-\frac12\right)^2+\left(y-\frac12\right)^2=\frac12$, a circle through $O$, $A=(1,0)$ and $(0,1)$; by symmetry the other quadrants are similar. So $OA=1$ and $AB=\sqrt2$, where $B=(0,1)$.
The region consists of the square with vertices $(\pm1,0),(0,\pm1)$ (side $\sqrt2$, area $2$) and four semicircles of radius $\frac{\sqrt2}2$ on its sides, i.e. $2$ circles. The area is $2+2\cdot \pi \cdot \frac12=2+\pi$.
