# Radical Axis of 2 Circles

Radical Axis of 2 Circles:Given two circles \(\Gamma\) and \(\Psi\), the radical axis of the two circles is the set of points with equal power with respect to both circles. \(_\square\)

Recall that the power of a point \(P\) with respect to the circle with center \(O\) and radius \(r\) is given by \(\lvert\overline{PO}\rvert^2-r^2\).

It turns out that the radical axis is always a line. This is easily proved using Cartesian coordinates. The proof is as follows:

## Prove that the radical axis of two circles is a line.

On the Cartesian plane, suppose the two circles \(\Gamma\) and \(\Psi\) have radii \(r\) and \(s\), and centers \(R(a,0)\) and \(S(b,0)\), respectively, where \(a\neq b\) (notice that we've taken their centers on the \(x\)-axis, because we can always set up a coordinate system with the two centers lying on an axis). Let \(\mathcal E\) be the radical axis of the circles. Then for any point \(P(x,y)\) lying on \(\mathcal E\), we must have the equality \(\lvert\overline{PR}\rvert^2-r^2=\lvert\overline{PS}\rvert^2-s^2\) since the powers are equal. Applying the Pythagorean theorem (or the distance formula), we find \(\lvert\overline{PR}\rvert^2=(x-a)^2+y^2\) and \(\lvert\overline{PS}\rvert^2=(x-b)^2+y^2\). So the equation becomes

\[(x-a)^2+y^2-r^2=(x-b)^2+y^2-s^2~\Longleftrightarrow ~x = \dfrac{a+b}{2} - \dfrac{r^2-s^2}{2(a-b)}.\]

Notice that \(a,b,r,s\) are constants. So we get two things: firstly, since the final equation is linear, the radical axis is a line; secondly, since we got a solution in \(x\), the radical axis is parallel to the \(y\)-axis, thus perpendicular to the \(x\)-axis. But the \(x\)-axis is the line joining the centers of the two circles.

\(_\square\)Hence, the radical axis is a line perpendicular to the line joining the centers of the two circles.

Applying the above result, we can notice another interesting thing. Note that points on the circumference of any circle has power \(0\) with respect to that circle, since then \(\lvert\overline{PO}\rvert=r\) and so \(\lvert\overline{PO}\rvert^2-r^2=0\). So, if two circles intersect at two distinct points (that is, not tangent to each other), then their radical axis is the line joining those intersections. Because the intersecting points have equal power (zero) with respect to both circles, so they must lie on the radical axis. But since radical axis is a line, that line must be the one joining these two points!

Similarly, if the circles are tangential, then the radical axis is the common tangent of the circles at the touch point, because the touch point lies on the radical axis and the tangent is perpendicular to the line joining the centers.

**Cite as:**Radical Axis of 2 Circles.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/radical-axis-of-2-circles/