The question goes like:
Find the angle of intersection of 2 circles S:$x^2+y^2-4x+6y+11=0$ and S': $x^2+y^2-2x+8y+13=0$...
I used the formula $$\cos x = \frac{\vert r_1^2+r_2^2-d^2\vert}{2r_1r_2}$$ Where $r_1,\;r_2$ are the radii of the circles and $d$ is the distance between the centers. I got the answer as $\cos x = \left|\frac{1}{\sqrt 2}\right|$ which is $45°$. But the answer sheet says $135°$. Should I take the positive or negative value of cos here? How do I know which one to take in general?
The angle between the two circles is the angle formed by the tangents in an intersection point. Look here and the image below
hope this helps
$$...$$