The circles $x^2+y^2+x+y=0$ and $x^2+y^2+x-y=0$ intersect at an angle of 

The angle 8 of intersection of two cirlces is given by 

$\cos \theta =\frac{r_1^2+r_2^2-C_1{C}_2^2}{2r_1{r}_2}$

where $r_1 , r_2$ are radii of two circles and $C_1 C_2$ is the distance between their centres. 

Here, $r_1=\sqrt{\frac{1}{4}+\frac{1}{4}}=\sqrt{\frac{1}{2}}=r_2$ and $C_1{C}_2=1$

$\therefore \cos \theta =0\Rightarrow \theta =\frac{\pi }{2}$

$\underline{ALITER}$ For the two circles, we find that $2\left(g_1{g}_2+f_1{f}_2\right)=c_1+c_2$

hold. So, the angle of intersection is a right angle.