Common Chord and the Radical Centre of Three Circles

The common chord is the radical axis. Suppose two circles $S=0$ and $S'=0$ actually intersect at two distinct points $P,Q$ (rather than just having some abstract locus of equal power). Both $P$ and $Q$ lie on $S=0$ and on $S'=0$, so both have power $0$ with respect to each circle — in particular, equal power with respect to both. That means $P,Q$ both lie on the radical axis $S-S'=0$. Since a line is determined by two points, the radical axis of two intersecting circles is nothing but the line through their common chord — the equation of the common chord is simply $S-S'=0$. (If instead the two circles merely touch at one point, the same equation $S-S'=0$ gives their common tangent at that point — the "chord" has shrunk to a point of contact, and the perpendicular-to-line-of-centres property forces the tangent line, since a common tangent at a point of contact is always perpendicular to the line of centres there.)

Finding the length of a common chord. Once you have the chord's equation, its length is a short right-triangle computation: find the perpendicular distance $p$ from either circle's centre to the chord, then (half-chord)$^2$ + $p^2$ = radius$^2$, so half-chord $=\sqrt{r^2-p^2}$ and the full chord length is $2\sqrt{r^2-p^2}$.

Worked example. Find the common chord and its length for $S\equiv x^2+y^2-2x-4y-4=0$ and $S'\equiv x^2+y^2-6x-4y+4=0$.

$S-S'=0$: $(-2x-4y-4)-(-6x-4y+4)=0 \Rightarrow 4x-8=0 \Rightarrow x=2$ — a vertical line.

Circle $S$ has centre $(1,2)$ and radius $\sqrt{1+4+4}=3$. The perpendicular distance from $(1,2)$ to the line $x=2$ is $|2-1|=1$. So half-chord $=\sqrt{3^2-1^2}=\sqrt8=2\sqrt2$, and the common chord has length $4\sqrt2$.

The radical centre of three circles. Take three circles whose centres are not collinear. Pairing them up gives three radical axes: one for ($S_1,S_2$), one for ($S_2,S_3$), one for ($S_3,S_1$). Writing each as $2(\cdot)x+2(\cdot)y+(\cdot)=0$ and adding all three equations together, the coefficients telescope to exactly $0=0$ — which is the algebraic signature of three lines being concurrent (any one is a linear combination of the other two). So all three radical axes meet at a single point, called the radical centre of the three circles.

A useful bonus fact: since the radical centre has equal power with respect to all three circles, the tangent lengths from the radical centre to all three circles are equal. This is exactly the property used to build a circle orthogonal to three given circles at once: centre it at the radical centre, and give it radius equal to that common tangent length — it will then cut all three circles at right angles (by the $d^2=r_1^2+r_2^2$ picture from §2.3, applied three times).

Worked example. Find the radical centre of

and verify the equal-tangent-length property. …