The Radical Axis of Two Circles

Power of a point, recalled. For a circle $S\equiv x^2+y^2+2gx+2fy+c=0$ and any point $P(x_1,y_1)$, the quantity $S_{11}=x_1^2+y_1^2+2gx_1+2fy_1+c$ is called the power of $P$ with respect to the circle. It is positive if $P$ is outside the circle (and then $\sqrt{S_{11}}$ is the length of a tangent from $P$), zero if $P$ is on the circle, and negative if $P$ is inside.

Definition. Given two circles $S=0$ and $S'=0$, the radical axis is the locus of points whose power with respect to $S$ equals their power with respect to $S'$.

Deriving its equation. Let $S\equiv x^2+y^2+2gx+2fy+c=0$ and $S'\equiv x^2+y^2+2g'x+2f'y+c'=0$, and let $P(x_1,y_1)$ be a point with equal powers:

The $x_1^2$ and $y_1^2$ terms cancel — that's the key structural fact, since both circles have the same leading coefficient (both are $x^2+y^2+\dots$, coefficient $1$). What's left is linear:

So the radical axis is a genuine straight line, with equation simply $S - S' = 0$ — subtract one circle's equation from the other's and the quadratic parts vanish automatically. (For this to be a line and not a triviality, the circles must be non-concentric, i.e. $(g,f)\ne(g',f')$; two concentric circles of different radii have no point of equal power at all, and if their radii happen to be equal every point qualifies — but that's just two names for the same circle.)

Key property — it's perpendicular to the line of centres. The radical axis's slope works out to $-\dfrac{g-g'}{f-f'}$, while the slope of the segment joining the two centres $(-g,-f)$ and $(-g',-f')$ is $\dfrac{f-f'}{g-g'}$. Multiplying these two slopes gives $-1$, so the radical axis is always perpendicular to the line joining the two centres — a fact worth remembering, because it's often the fastest way to sanity-check a radical-axis computation.

Worked example. Find the radical axis of $S\equiv x^2+y^2-4x+2y-4=0$ and $S'\equiv x^2+y^2+2x-6y+2=0$, and verify the perpendicularity property.

Solution. $S-S'=0$ gives

$(-4x+2y-4)-(2x-6y+2)=0 \ \Longrightarrow\ -6x+8y-6=0 \ \Longrightarrow\ 3x-4y+3=0.$ …