Case 1: $x,y>0$ We have $x^2+y^2\le4x+4y\Leftrightarrow (x-2)^2+(y-2)^2\le8$, which is a circle with center $(2,2)$ and radius $2\sqrt2$. The area is an isosceles right triangle with a semicircle, which has area $\frac124^2+\frac12\pi\cdot8=8+4\pi$. What we have so far: Case 2: $x>0,y<0$ We have $x^2+y^2\le4x-4y\Leftrightarrow (x-2)^2+(y+2)^2\le8$, which is a circle with center $(2,-2)$ and radius $2\sqrt2$. The area is an isosceles right triangle with a semicircle, which has area $\frac124^2+\frac12\pi\cdot8=8+4\pi$. What we have so far: Case 3: $x<0,y<0$ We have $x^2+y^2\le-4x-4y\Leftrightarrow (x+2)^2+(y+2)^2\le8$, which is a circle with center $(-2,-2)$ and radius $2\sqrt2$. The area is an isosceles right triangle with a semicircle, which has area $\frac124^2+\frac12\pi\cdot8=8+4\pi$. What we have so far: Case 4: $x<0,y>0$ We have $x^2+y^2\le-4x+4y\Leftrightarrow (x+2)^2+(y-2)^2\le8$, which is a circle with center $(-2,2)$ and radius $2\sqrt2$. The area is an isosceles right triangle with a semicircle, which has area $\frac124^2+\frac12\pi\cdot8=8+4\pi$. The final figure has an area of $32+16\pi$. Final picture: Happy belated St. Patrick's day!