## 1. Radical axis of two circles

It is the locus (line r) of points from which equal tangents to the two circles {a,b} can be drawn. It coincides with the locus of centers of circles simultaneously orthogonal to two given circles. Concentric circles have no real radical axis (their radical axis coincides with the line at infinity).

Each point D on the radical axis defines four other circles tangent simultaneously to the given circles a and b, at the points of tangency of the tangents from D. This can be read also the other way around: When a circle c touches simultaneously two other circles {a,b} then the common tangents at the tangency points concur at the radical axis of {a,b}. The calculation below gives the distances |AC|, |CB| and their ratio.
|AC| + |CB| = d,
AC² - CB² = AD² -DB² = r1² - r2².
d1 = |AC| = (d²+ r1²-r2²)/(2*d),
d2 = |CB| = (d²+ r2²-r1²)/(2*d).
AC/CB = (d²+ r1²-r2²)/(d²+ r2²-r1²) = (1+k)/(1-k), where
k = (r1² - r2²)/d².

## 2. Common tangents middles

This is a consequence of the definition. The middles of the four common tangents of two circles exterior to each other are on the radical axis of the two circles, hence they are collinear.

## 3. Construction of the radical axis

Given two non-concentric circles c1(O1,r1), c2(O2,r2), to construct their radical axis draw an arbitrary circle (c) intersecting {c1, c2} along lines {L1, L2} respectively. Project the intersection point A of {L1, L2} to point B on line O1O2. The radical axis is line AB.

## 4. A property of the radical axis

Given two points A,B and a circle c(O,r) the radical axis Ld of circle (c) and an arbitrary circle d(O',r') passing through points {A,B} passes through a fixed point D on line AB.

In fact, D on the radical axis means : (DA*DB) = (DI*DJ) =>
(DF + FA)*(DF - FA) = (DO + OJ)*(DO - OJ) =>
DF2 - FA2 = (DO2) - OJ2 = (DE2+EO2) - OJ2 =>
DF2 - DE2 = FA2 + EO2 - OJ2.
The right side being constant and points F, E being also constant implies that D is constant.

## 5. Application to an Apollonius problem

To construct a circle (d) passing through two given points {A,B} and tangent to a given circle c(O,r).

The desired circle will belong to the bundle of circles passing through {A,B}, thus the radical axis of it and c will pass from the known point D. If Q is the contact point of c and d then the tangent there, for the same reason, will pass through D. Thus, to construct (d) locate first D and draw tangents to c(O,r) from there.

To construct D draw an arbitrary circle (d0) through {A,B} and intersecting (c) at a point C. D is the intersection point of the common chord CD of the two circles {d0, c}.
From D draw tangents to c and consider the contact points {E,E'}. The desired circles are (ABE) and (ABE').
There are either two solutions, when both {A,B} are inside/outside c or none if this condition does not hold.

## 5. Radical axis and homothety centers

Given two circles c1(O1,r1), c2(O2,r2) and a point D on their radical axis, draw tangents DA, DB, DC, ... to the circles. Lines of contact points AB, AC, ... pass through one of the similitude centers I, J of the two circles.

To see it in one case, for line AC say, consider the other intersection point A' of c1 with line AC and the tangent there intersecting AD at A*. A*A = A*A' and DA = DC by the property of the radical axis. Thus triangles A*AA' and DAC are isosceli and A*A is parallel to DC, hence O1A' and O2C are parallel. This implies that AC passes through the similarity center characterized by the equation JO1/JO2 = r1/r2.