The angle of intersection of the circles $x^2+y^2=4$ and $x^2+y^2=2x+2y$, is

We know that the angle of intersection of two circles of radii $r_1$ and $r_2$ is given by

$\cos \theta =\frac{d^2-{r_1}^2-{r^2}_2}{2r_1{r}_2}$,

where d is the distance between  their centres

Here, $r_1=2,r_2=\sqrt{2}$ and $d=\sqrt{2}$.

$\therefore \cos \theta =\frac{2-4-2}{2\times 2\times \sqrt{2}}=\frac{-1}{\sqrt{2}}\Rightarrow \theta =\frac{3\pi }{4}$