Hubert Shutrick

In the article Gergonne's Solution to Apollonius' Problem, it is shown how two of the circles that are tangent to three given circles are constructed. In the case when the discs of the three circles do not intersect one another, there are eight touching circles and here it will be shown how the other six can be found using Gergonne's method.

Let the given circles be a(A), b(B), c(C) , where A,B,C are their centres and let r(R), s(S) be the two tangential circles such that s includes a but not b and c whereas r includes b and c but not a.

As in the article above, The first step is to construct the circle o(O) that is orthogonal to the three given circles since inversion with respect to it interchanges r and s and, since these two intersect in this case, it is very clear that pairs from o , r and s share the same radical axis.

Let M on AB between A and B be the intersection of the common tangents to the two circles and, similarly, N on AC between A and C. The points of contact between the given circles and the ones to be constructed can be obtained as follows: if the line of intersection of a and o meets MN at P, then the two tangents from P to a have contact points T_{ar} where r touches and T_{as} where s touches. Similarly, the contact point $ T_{br} $} can be found and so the centre of r is the intersection of AT_{ar} and BT_{br} .

The reason why this works depends on the fact that it will be shown that MN is the radical axis of the circles r and o and, since the radical axis of a and r is the tangent to a at T_{ar} , it must intersect MN where the radical axis of a and o does.

Proof

Let M' be the intersection of the chords T_{ar}T_{br} and T_{as}T_{bs} . Consider the triangle T_{ar}T_{as}M' . The segment T_{ar}T_{as} is the base of an isosceles triangle with vertex A so let its base angles be \alpha . Using this notation, \angle T_{ar}T_{as}M' = \sigma - \alpha and \angle T_{as}T_{ar}M' = \pi -\alpha - \rho so \angle T_{as}M'T_{ar} = 2\alpha + \rho - \sigma . From the other triangle, \angle T_{bs}M'T_{br}= 2\beta + \sigma - \rho giving that \alpha + \rho = \beta + \sigma and the four tangential points are concyclic. The chords T_{ar}T_{br} and T_{as}T_{bs} are the intersecting chords of this circle with r and s so M' is on the common radical axis of o , r and s.


Consider now the similarity transformation that first multiplies vectors radiating from M' with the scalar - M'T_{br} / M'T_{as} and then reflects in the bisector of \angle T_{br} M'T_{bs} . It takes the triangle T_{ar}T_{as}M' to T_{bs}T_{br}M' and therefore the circle a to b while leaving M' fixed so M' = M. In the same way, N will be on the common radical axis of o , r and s, which must be MN as required.