Find the angle of intersection of circles $x^2+y^2-6x+4=0 \ \&\ x^2+y^2-2x-2y-8=0$ Find the angle of  intersection of circles $$x^2+y^2-6x+4=0 \\  x^2+y^2-2x-2y-8=0$$
my answer is : 41.14 degrees. but i'm not sure if it's right. please help me.

Write the equations as
$$
(x-3)^2+y^2=5
$$
and
$$
(x-1)^2+(y-1)^2=10
$$
Thus, we have a triangle with sides $\sqrt{5}$, $\sqrt{10}$, $\sqrt{5}$:
$\hspace{3cm}$
The Law of Cosines says
$$
5=5+10-2\sqrt{5}\sqrt{10}\cos(\alpha)
$$
which implies
$$
\cos(\alpha)=\frac1{\sqrt2}
$$
We could also recognize the $45{-}45{-}90$ triangle.
