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.

If the result is true regardless of the other measurements than b, then we can assume that the radius, k, of the circle of inversion, is 1.

So that we can make calculations let's take the distance of the line from the center of the circle of inversion to be d, and the given circle has radius a. Let's assume, again because we are assured it doesn't matter, that the given point and the center of the given circle are all on a line through the center of the circle of inversion and perpendicular to the given line; call this line the axis.

The line maps into a circle going through the center of the circle of inversion and through a point on the axis 1/d units from that center.

The given circle's intersections with the axis lie at d+b+a and d+b-a units from the center of inversion and so their images, at diametrically opposite points on the image of the circle, are at 1/(d+b+a) and 1/(d+b-a). The center of this image, which then is the image of the given point, is therefore at the average of these two distances, or (1/(d+b+a) + 1/(d+b-a)) / 2. It's original point is therefore at 2 / (1/(d+b+a) + 1/(d+b-a)).  Since the line is d units away from the center of inversion, the point is 2 / (1/(d+b+a) + 1/(d+b-a)) - d units away from the line.

Plugging numbers in, it does not seem that the result is independent of d as implied by the question. It would seem that not enough information has been supplied. It would also seem to depend upon k as well, which we had assumed here was 1.

Edited on February 22, 2006, 10:32 am

Posted by Charlie on 2006-02-22 10:25:11