The Problem of Apollonius - cases involving point-circles

Suppose that each of L, M and N is either an i-line or a point.
We say that an i-line T touches a point P if P lies on T.
The problem is to decide how many i-lines touch L, M and N.

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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