A circle (of radius a), a line, and a point are mapped by inversion into two concentric circles and the center of those concentric circles. If the distance from the given circle's center to the line is b, then what is the distance from the point to the line?

Inversion Defined:

Let O be the center of a circle of radius k. An inversion with respect to circle O is a mapping f:R² -> R² such that for all P in R² (not O), P' = f(P) lies on ray OP and
|OP'||OP| = k².

See www.geocities.com/bractals/inv.jpg

for graphical description of inversion.

The answer is [-k+sqrt(k^2+8kb^2)]/4b

So the line must pass inside the circle O (with center "o") - let's call the midpoint of the secant "x". It maps to a circle with one end of its diameter at o and the other twice as far as the center of the concentric circles (let's call the center of the concentric circles "c", and the far point "y").  The point p is at the center of a secant of a parallel line which also passes inside O and maps to a circle with one end of its diameter at o and the other at c.  The line segment connecting c to x (which has distance b) maps to the line segment from p to y. This distance is equal to 2oc-op.  So my collinear points in order are o,x,p,c,y. We are solving for px. xc=b oy=(k/ox) oc=k/2ox. op maps to yc which equals oc so k/op = k/2ox meaning op=2ox. and ox=px.

Together this gives 2px + k/b = k/px

Converting to a quadratic equation in terms of px we have:

2b(px)^2 + k(px) -kb = 0

Solving for px (which must be positive)

px = [-k + sqrt(k^2 + 8kb^2)]/4b


I admit that I think my likelihood of error here is high.

Posted by Eric on 2006-02-21 18:25:24