Write the equation as $x^2+y^2=k(|x|+|y|)$ with $k=2$. In the first quadrant it is $\left(x-\frac k2\right)^2+\left(y-\frac k2\right)^2=\frac{k^2}2$, a circle through the origin and through $(k,0)$ and $(0,k)$; the segment joining the last two points is a diameter. So the part of the region in the first quadrant is the right triangle with vertices $(0,0),(k,0),(0,k)$ of area $\frac{k^2}2$ together with a semicircle of radius $\frac k{\sqrt2}$ of area $\frac{\pi k^2}4$. By symmetry the total area is $4\left(\frac{k^2}2+\frac{\pi k^2}4\right)=k^2(2+\pi)$.
