Without loss of generality, we can place P1 at (0,0) and P2 at (1,0). Any other distance can be scaled as needed.
For simplicity, let the ratio of the squares of the distances be k. We'll also let k=c^ so that c is the ratio of the distances.
Then
x^2 + y^2 = k ((x-1)^2 + y^2) = kx^2 - 2kx + k +ky^2
(1-k)x^2 + 2kx + (1-k)y^2 = k
This is certainly a conic, and is in fact a circle, as it has two real intersections with the x-axis, as when y = 0, x = (-2k +- sqrt(4k^2 + 4(k-k^2))) / (2(1-k)) = 2k +- 2 sqrt(k) / (2(k-1)) = (k +- sqrt(k))/(k-1), so if k=c^2, where c is the ratio of the distances, as k was the ratio of the squares of the distances, that's (c^2 +- c) / (c^2-1).
So the center of the circle is at c^2 / (c^2 - 1) and the radius is c/(c^2-1). Remember of course to scale up by the distance from P1 to P2.
For example with c=2, and our (0,0) and (1,0) original points, the center is at 4/3 and the radius is 2/3, so indeed (2/3,0) is twice as far from (0,0) as from (1,0) and (2,0) is also twice as far from (0,0) as from (1,0).
A little research shows that this is one of the definitions of Apollonian circles. There are others, but unfortunately there doesn't seem to be a name particular to this type of circle, not shared with other definitions.
It's also one of the bases of binaural hearing. If not for the distortion caused by the shapes of our outer ears, location of sounds in the space around us would be in such an Apollonian sphere based upon the ratio of the loudness heard in each ear. Combined with the difference in time that the sound reaches each ear, resulting one nappe of a hyperboloid, that would point to a circle in space (the intersection of the sphere with the hyperboloid) from which the sound could be originating. All these shapes are distorted by that external ear, but the working is similar.
Edited on December 7, 2006, 10:50 am
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Posted by Charlie
on 2006-12-07 10:48:33 |
solution
