The Problem of Apollonius

We are familiar with the fact that, given any triangle, we can draw a circle inside the triangle, touching all three sides.
This is the incircle, its centre is the point where the internal angle bisectors meet. Indeed, this result and its proof are
valid in hyperbolic as well as euclidean geometry.
If we look at the lines through the vertices rather than just the segments, we can draw three further circles, each
touching all three lines. These are the excircles. They can be found by using two external bisectors and one internal.

These are illustruated in the CabriJava applets below. You can move the vertices A, B, C to verify the results.

Apollonius asked what happens if we replace some of the lines with circles. In the language of inversive geometry,
we might ask the following questions:
Given three distinct i-lines, can we find an i-line touching all three?
If so, how many such i-lines are there?

The problem is often extended by allowing "point-circles" - i.e. a point may be regarded as a "circle" of zreo radius.
These do not fit neatly into the inversive description. We shall deal with this extension separately.

We shall see that:
There are many configurations where there are no such i-lines.
There is one type where there are an infinite number of i-lines.
Otherwise, the number may be 2, 4, 6 or 8.

The euclidean result mentioned about says the if we have three non-parallel extended lines, then there are four
circles which touch all three. Observe that two extended lines meet at ∞, so 'touch' if and only if this is the only
intersection, i.e. the corresponding lines are parallel. Since we have non-parallel lines, there is no extended line
touching all three. Thus, there are exactly four i-lines in this case.

The euclidean result

The incircle
One of the three excircles

general results