Hubert Shutrick

Apollonius worked on the problem of finding a straight edge and compass construction for a circle tangential to three given circles. This note describes how these constructions can be done for all configurations. It was written before the comprehensive posting appeared.

First, a method of solution in the case when two of the circles intersect. It is based on the easy construction in the special case of finding a circle that is tangent to two given lines and a given circle, that is, if, for each line, a parallel tangent to the circle is constructed and the line joining their intersection to the intersection of the given lines cuts the circle at a point S, then there is one of the required circles touching at S.

    

This construction works because the existence of the required circle implies that the homothety that takes it to the given circle c(BDLK) projecting through the intersection point S takes BD to a parallel tangent of c which is well-determined without knowing where S lies. Similarly, the image of LK is determined and their intersection R must be the image of J, the intersection of BD and LK. Hence, S is where RJ meets the circle c.

Elsewhere it is pointed out that the general case can be reduced to this special case by inverting with respect to a circle centred at one of the intersection points, constructing the circle in the special case and then inverting back. It should work well when all the circles intersect pairwise if the inversion circle is chosen to be the one orthogonal to the given circle that doesn't pass through the chosen centre of inversion. In that case. the given circle inverts to itself and the inverts of the intersection points are obvious. To extend it cases when these intersection points don't exist, one can find another larger circle orthogonal to the inversion circle that does have the required intersection points.

Here the method is to imitate the special construction without transforming. Imagine that a sphere has been placed on the plane and the three circles are projected onto it by stereographic projection, after which the sphere is rolled so that the image of one of the intersection points of the two circles is the top point and then projected down again. It should then resemble the diagram above. We now imitate the construction above for the three circles remembering that lines above correspond to circles. The circle corresponding to the line AG above is determined as being orthogonal to the two circles involved since the centre A is not preserved. I

   

In the diagram, the given circles are c(BDLK) centre A, d(JBDE) centre C and k(ELKJ) centre I. We construct the Apollonian circle in the arc triangle JBK so E corresponds to the point at infinity. First, construct the circle f(EG) whose centre is F, the intersection of the perpendicular to CE at E and BD, because it is orthogonal to c and d and because G on the circle c will be the point where a circle h(EGR) is tangent to c and will correspond to the tangent line in the special case. The centre of h is H the intersection of CE and AG. In the same way, the circle p(EO) centre N leads to r(EOR) centre Q corresponding to the other tangent. The two circles h and r have the common point R and the circle s(ERJ), whose centre is so far away that it looks straight, meets the arc KB of c at the tangency point S for the required circle. The circles t(EGS) and u(EOS) give the other two tangency points U and T.

If the points B and D had not existed because the two circles had not intersected, then the centre F would have been the intersection of the tangent with their radical axis.

It is interesting to note that it would be impossible to do this construction accurately enough using a straight edge and compass. The tools used in the Geogebra diagram do just what the ruler and compass should do but with sufficient accuracy.

Another Method

For the configuration where two of the circles have one intersection inside the third circle and one outside, there is also a reasonably easy solution based on the lemma of Sawayama.

   

Let o(_) denote the circle through the points inside the brackets and let O(_) denote its center. The point D is inside o(ABC) and the other two are o(BDCN) and o(ADN) where N is outside o(ABC). Bisect the angle ∠O(ABN)AO(ACN) and find its intersection with the midpoint normal of the segment AN since it will be the centre of a circle o(AIN), which corresponds to the bisector of the angle A in the lemma. Construct the circle corresponding to the bisector of angle B in the same way and let I be their other intersection since it corresponds to the incentre of the triangle. Construct the circle o(DN) whose centre is on the bisector of ∠O(ADN)DO(BDN) and the circle o(IFEN) that is orthogonal to it, the points F and E being its intersection with o(ADN) and o(BDCN) respectively. These points are the points where the required circle touches o(ADN) and o(BDCN) since it will also be tangent to o(ABC).

The point I in the construction, corresponding to the incentre of the triangle ABC in the lemma, does determine the circle that is tangent to the arcs BC, CA and AB in diagram. It is not the centre of this circle but the circles through N and I that are orthogonal to the arcs intersect them where the circle touches them. Thus it gives a solution to the problem when the three given circles are concurrent.

If the plane is considered as the Argand diagram of complex numbers, the transformation between the special case and the more general case is an example of a Möbius transformation. Its salient properties are that it is conformal, i.e. preserves angles, and that it takes circles/lines to circles/lines.

The other Case

If the circles a, b and c do not intersect, the methods above are not applicable but there is the definitive solution by Gergonne. Here we suggest a GeoGebra construction that is not ruler and compass but can be reasonably accurate. If one of the eight Apollonian circles touches the three given circles at points R, S and T, then the tangents at these points are its radical axes with the given circles and so form a triangle whose vertices lie on the three radical axes ca, ab and bc of the given circles.

We construct a mapping from ca to itself as follows: for a point P on ca let a tangent from it to a intersect ab at U; then a tangent from U to b intersects bc at V; finally a tangent from V to c intersects ca at the image point. Sliding P along ca should find where P is mapped to itself and the accuracy can be improved by zooming and correcting.

Since there are two choices of tangent at each stage, all eight circles can be obtained this way.

Two Circles and a Line

The degenerate case to find a circle tangent to two given circles and a given line has a similar construction.

   

The existence of the required circle implies that the homotheties from it to the given circles b and c through the common points take the line to parallel tangents whose points of contact S and T are therefore known. The common tangents of the required circle with b and c intersect on their radical axis. Therefore, from a point U on the radical axis, construct the tangent to b at Q and let QS intersect the line at P. Then let PT intersect c at R and the tangent at R intersect the radical axis at V. If V coincides with U then the circle through PQR solves the problem.