Answer: B

Solution:

By symmetry, consider the first quadrant, where the equation is $x^2+y^2=x+y,$ or $\left(x-\tfrac12\right)^2+\left(y-\tfrac12\right)^2=\tfrac12.$ This is a circle centered at $\left(\tfrac12,\tfrac12\right)$ passing through $(1,0)$ and $(0,1);$ since the center is the midpoint of that chord, the enclosed first-quadrant region is the right triangle with legs to $(1,0)$ and $(0,1)$ (area $\tfrac12$) plus a semicircle of radius $\dfrac{\sqrt2}{2}$ (area $\dfrac\pi4$). Multiplying by $4$ for all quadrants gives $4\left(\tfrac12+\tfrac\pi4\right)=\pi+2.$

Thus, the correct answer is B.