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.

In fact, D on the radical axis means : (DA*DB) = (DI*DJ) =>

(DF + FA)*(DF - FA) = (DO + OJ)*(DO - OJ) =>

DF

DF

The right side being constant and points F, E being also constant implies that D is constant.

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 (d

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.

To see it in one case, for line AC say, consider the other intersection point A' of c

ThreeCirclesProblem.html

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