Angle between Two Intersecting Circles

When two circles intersect at a point $P$, the angle between the circles at $P$ is defined as the angle between their two tangent lines at $P$ -- and since each tangent is perpendicular to the radius drawn to the point of contact, this is the same as the angle between the two radii $C_1P$ and $C_2P$, where $C_1,C_2$ are the two centres. (This angle is the same at either point of intersection, by the symmetry of the configuration, so 'the' angle between two circles is well defined.)

Let the circles have centres $C_1,C_2$, radii $r_1,r_2$, and centre-to-centre distance $d=C_1C_2$. Triangle $C_1PC_2$ has sides $r_1,r_2,d$ and included angle $\theta$ (the angle between the circles) at the vertex $P$. The cosine rule in this triangle gives immediately

For circles given in the general form $S:x^2+y^2+2gx+2fy+c=0$ and $S':x^2+y^2+2g'x+2f'y+c'=0$, the centres are $(-g,-f)$ and $(-g',-f')$ and the radii satisfy $r_1^2=g^2+f^2-c$, $r_2^2=g'^2+f'^2-c'$. Substituting $d^2=(g-g')^2+(f-f')^2$ into the boxed formula and expanding, every squared term ($g^2,g'^2,f^2,f'^2$) cancels, leaving the purely coefficient-based form

This lets the angle between two circles be found directly from their equations, without separately writing down the centres and radii, though doing so first is often clearer.

Subtracting the two equations, $S-S'=0$, cancels the $x^2$ and $y^2$ terms and leaves a linear equation -- the common chord of the two circles when they meet in two real points (every common point of $S=0,S'=0$ satisfies this linear equation too), or their common tangent at the single point of contact if the circles merely touch.