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Inversion Distance (Posted on 2006-02-21) Difficulty: 3 of 5
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.

See The Solution Submitted by Bractals    
Rating: 2.8000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): problem resolved | Comment 11 of 32 |
(In reply to re: problem by goFish)

You are right, I was getting ahead of myself and not being clear - dumb of me.

I was assuming the given circle was mapped to itself. (If k is variable in the result this should hold in generality.)

My result was derived from k^2 = 2sb + 2s^2 where s is the distance from the point to the line (also the line to the origin).

Because of how I have chosen my k, my circle intersects the inversion circle at a right angle so by Pyth Thm: a^2 + k^2 = (b+s)^2 or  k^2 = (b+s)^2 - a^2 which equals  2sb + 2s^2

So: (b+s)^2 - a^2 = 2sb + 2s^2 or b^2 - a^2 = s^2

or s = sqrt(b^2 - a^2)

Okay that looks a lot better, is independent of k, and makes sense that b>a otherwise the circles couldn't be concentric.


  Posted by Eric on 2006-02-22 17:08:39
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