Apollonian Circles Theorem
Given two points A and B and a number r. What is the locus of points P such that AP/BP = r? The answer is a circle. The circle which is known as the Apollonian Circle. For every positive r there is a different one.
This problem has been treated elsewhere. Here we present a different solution based on the inversion transform. Along the way we show that the whole family of Apollonian circles can be inverted into a family of concentric circles.
Solution
Copyright © 1996-2008 Alexander Bogomolny
Apollonian Circles Theorem
Consider a circle of radius R centered at A. The calculations are simplified by taking R = 1, but, in principle, any circle with center at A will do. So, assume R = 1. Let t be the inversion in this circle. Set t(B) = B' and t(P) = P'.
This means, in particular, that
Triangles AB'P' and APB, in which the sides satisfy (1), also share the angle at A. They are, therefore, similar. It then follows that
From here,
Combining (1) and (3) gives
| (4) |
| B'P' | = BP / (AP·AB) |
| | = 1/rAB. |
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This tells us that P' lies on a circle of radius 1/rAB centered at B'. In other words, the inversive image of the locus of points P is a circle centered at B'. To repeat, the Apollonian circle defined by the point circles A and B and r > 0 is mapped by t to a circle centered at t(B). This is true for any Apollonian circle defined by A and B, so that the whole family of them is mapped on the family of concentric circles with center t(B).
In establishing Steiner's Porism we showed that any two non-intersecting circles can be inverted into a pair of concentric circles. The above strengthens this assertion with a more direct proof.
The Apollonian circles defined by two point circles are said to be coaxal. This is one of the three varieties of coaxal families.
References
- D. A. Brannan et al, Geometry, Cambridge University Press, 2002
Copyright © 1996-2008 Alexander Bogomolny
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