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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)

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Solution Algebraic solution | Comment 13 of 32 |

I initially tried drawing out the circles and lines, when I realized that a much simpler way would be to reduce to a single dimension.  That is, I will only consider the locations where the circles, line and dot cross the line going through the diameters of all the circles.

Without loss of generality, assume the center of the circle with respect to which we invert is at 0.
Let k = the radius of the circle with respect to which we invert
Let x = the location of the line before inverting
Let c = the location of the center of the circle before inverting
Let p = the location of the point before inverting
Let r = the radius of the circle before inverting
Now let's see how many of these variables cancel out...

As given,
c = x + b

Since the circles after inverting are concentric,
k/p = (k/(c+r) + k/(c-r))/2 = k/2x
p = 2x
k/2x = kc/(c^2-r^2)
2xc = c^2 - r^2

Using the given,
2x^2 + 2bx = x^2 + 2bx + b^2 - r^2
x^2 = b^2 - r^2
x = sqrt( b^2 - r^2 )
|x - p| = x = sqrt( b^2 - r^2 )

I'm surprised that r doesn't cancel.  This makes me suspect error, or perhaps that I am not yet finished.

Edit: changed according to goFish's correction.  Thanks!

Edited on February 23, 2006, 7:50 pm
  Posted by Tristan on 2006-02-22 21:58:05

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