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

Solution :
Equations of the circles are
$$\eqalign{ & {x^2} + {y^2} = 4......\left( 1 \right) \cr & {\text{and }}{x^2} + {y^2} = 2x + 2y......\left( 2 \right) \cr} $$
Centre of $$\left( 1 \right)$$ is $${C_1} \equiv \left( {0,\,0} \right)\,;$$   Radius of $$\left( 1 \right) = {r_1} = 2\,;$$
Centre of $$\left( 2 \right)$$ is $${C_2} \equiv \left( {1,\,1} \right)\,;$$   Radius of $$\left( 2 \right) = {r_2} = \sqrt 2 $$
$$d = $$  distance between centers $$ = {C_1}{C_2} = \sqrt {1 + 1} = \sqrt 2 $$
If $$\theta $$ is the angle of intersection of two circles, then
$$\eqalign{ & \cos \,\theta = \frac{{r_1^2 + r_2^2 - {d^2}}}{{2{r_1}{r_2}}} = \frac{{{{\left( 2 \right)}^2} + {{\left( {\sqrt 2 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}}{{2.2\sqrt 2 }} = \frac{1}{{\sqrt 2 }} \cr & \therefore \,\theta = \frac{\pi }{4} \cr} $$