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.

(In reply to How do I explain this? by Eric)

I was messing up dividing everything by k instead of k^2. Okay, now I think I have it:

I am going to just lay everything out on a cartesian plane centering the inverse circle at the origin. All the significant points will lie on the positive x-axis beginning with the inverse circle intersecting the x-axis at k.

Let our given line be perpendicular to and intersect the x-axis at s. It is easy to see that (k^2)/2s is the center of our circles. This is where our point p maps to (i.e. (k^2)/p) so p=2s and the distance from p to s is also s. That means we are solving for s. But we already know that the distance from s to (k^2)/2s is b so s + b = (k^2)/2s 

Which is to say 2s^2 + 2bs - k^2 = 0

Using the quadratic equation (correctly this time, again noting s is positive)

s = [-2b + sqrt(4b^2 + 8k^2)]/4 or reducing:

s= [-b + sqrt(b^2 + 2k^2)]/2

In Mindy's special case where b = k this resolves to s = k[sqrt(3)-1]/2

Posted by Eric on 2006-02-22 02:03:52