Locus of Points in a Given Ratio to Two Points
Given two points A and B and a number r. What is the locus of points P such that AP/BP = r?
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In the applet, r is always rational and is presented as a fraction r = M/N.
Solution
Copyright © 1996-2008 Alexander Bogomolny
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.
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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.
The circle we just constructed is known as a Circle of Apollonius . The family of all such circles obtained for different values of r and fixed points A and B is known as the Apollonian Family of Circles defined by the points A and B. Note that one of these circles (corresponding to r = 1) is actually a straight line. It is often convenient to think of straight lines together with common circles as generalized circle s, circles of infinite and finite radii.
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 © 1996-2008 Alexander Bogomolny
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