The Problem of Apollonius - cases involving point-circles

As in the problem with three i-lines, we may invert to get a configuration where all the i-lines are circles.
We then have four basic cases, depending on the number of i-lines involved - 0, 1, 2 or 3. Of course, the
case of three i-lines has already been considered. The remaining cases each involve at least one point.

If N is a point, then we may apply a further inversion in a circle with centre N, so that we can assume
that N is the point ∞. Then any i-line "touching" this is an extended line. Since we assume that N does
not lie on the other "circles", L, M are still points or circles.

1. Three points.
   There is precisely one extended line through the points L and M.
2. Two points and one circle.
   Suppose that L is a circle, and M a point.
   We require extended lines through M touching L.
   If M lies inside L, then there are no such extended lines.
   If M lies outside L, then there are two extended lines - the tangents to L through M.
   Thus, we have either no common tangents to L, M, N or exactly two.
3. One point and two circles.
   Suppose that L and M are circles.
   We require (extended) lines tangent to both.
   
   1. One of L, M lies strictly inside the other.
      There can be no common tangents in such a case.
   2. L, M touch internally.
      The only common tangent is that at the point of contact.
   3. L, M touch externally.
      Then there are three common tangents (including that at the point of contact).
   4. L and M meet at two points.
      Then there are two common tangents.
   5. Each of L, M lies ouside the other.
      Then there are four common tangents.

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