Angle Between Two Intersecting Circles

Setting up the idea. Two circles intersect when the distance between their centres is neither too large (circles too far apart) nor too small (one swallowed inside the other) — precisely when $|r_1-r_2| < d < r_1+r_2$, where $d$ is the distance between centres and $r_1,r_2$ the radii.

At a point $P$ where the two circles cross, each circle has its own tangent line. The angle between the two circles at $P$ is defined as the angle between these two tangent lines at $P$. (If the circles meet at a second point $Q$ too, the angle there works out to be the same angle — the formula below doesn't depend on which intersection point you pick.)

A formula using centres and radii. Let $C_1, C_2$ be the centres, $d=C_1C_2$, and let $\theta$ be the angle between the circles at $P$. Draw the two tangent lines at $P$; each is perpendicular to its own circle's radius at $P$ (radius $\perp$ tangent, a fact you already know). If the two tangents meet the line $C_1C_2$ at points $T_1,T_2$, then in triangle $C_1PC_2$, careful angle-chasing around the right angles at $T_1$ and $T_2$ shows

Now apply the law of cosines to $\triangle C_1PC_2$, using $C_1P=r_1$, $C_2P=r_2$:

which rearranges to the working formula:

Turning this into coefficients. If the circles are $S\equiv x^2+y^2+2gx+2fy+c=0$ and $S'\equiv x^2+y^2+2g'x+2f'y+c'=0$, then $C_1=(-g,-f)$, $C_2=(-g',-f')$, $r_1=\sqrt{g^2+f^2-c}$, $r_2=\sqrt{g'^2+f'^2-c'}$, and $d^2=(g-g')^2+(f-f')^2$. Substituting and simplifying $d^2-r_1^2-r_2^2$ collapses nicely (the $g^2,f^2,g'^2,f'^2$ terms cancel against the ones hiding inside $r_1^2,r_2^2$), leaving:

This is the version you'll actually use in problems — you never need to compute $d$, $r_1$, $r_2$ separately; just read off $g,f,c,g',f',c'$ and plug in.

Worked example. Find the angle between the circles $S\equiv x^2+y^2=25$ and $S'\equiv x^2+y^2-14x+40=0$.

Solution. Writing $S$ in general form, $g=0,f=0,c=-25$; for $S'$, $g'=-7,f'=0,c'=40$. …