2016 AMC 10B Problem 21

The equation is symmetric in all four quadrants. In the first quadrant it becomes $$x^2+y^2=x+y,$$ or $$(x-\tfrac12)^2+(y-\tfrac12)^2=\tfrac12.$$ In the first quadrant, the enclosed region is the triangle under $x+y=1$, with area $\frac12$, plus a semicircle of radius $\sqrt{\frac{1}{2}}$, with area $\frac\pi4$. Multiplying by $4$, the total area is $$4\left(\frac12+\frac\pi4\right)=2+\pi.$$ Thus, the correct answer is B.