Given two points A and B and a number r. What is the locus of points P such that AP/BP = r?
In the applet, r is always rational and is presented as a fraction r = M/N.
Apollonian Circles Theorem
The locus is a circle, unless of course r = 1, in which case it's the perpendicular bisector of AB. The proof exploits the properties of angle bisectors: internal and external.
To see that (there is an additional proof), construct points M and N on the line AB such that AM/BM = AN/BN = r. For r
1, M and N always exist. Note that both M and N lie on the sought locus.
The circle at hand has MN as a diameter. Indeed, M serves as the feet of the internal bisector of triangle APB at apex P, PN serves as the external bisector. Therefore, PM
PN. Chords PM and PN are perpendicular and therefore define a 90o inscribed angle. The angle subtends a 180o arc, which means that MN is a diameter of the circle.
For any P on the circle, the internal and external bisectors of angle APB pass through (the fixed points) M and N.
Since, for the same A and B, each of the Apollonian circles correponds to a different r, no two Apollonian circles intersect. For 0 < r < 1, the circles are closer to A and surround it. For smaller values of r, they are increasingly close to B. For r > 1, the circles surround B and, as r grows become increasingly close to it. For this reason, points A and B are often considered as circles, now point circles, circles of radius 0.
Copyright © 1996-2009 Alexander Bogomolny