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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(3): Note to Bractals | Comment 25 of 32 |
(In reply to re(2): Note to Bractals by Mindrod)

 
Let O be the center of inversion and P the center
of the circle to be inverted. Let the line OP
intersect the circle at points A and B (AB a diameter
of the circle). Clearly, f(A)f(B) will be a diameter
of the inverted circle. WOLOG let O be the origin (0,0),
A (a,0), B (b,0), and P ((a+b)/2,0). So

    f(A) is (k^2/a,0)
    f(B) is (k^2/b,0)
    f(P) is (k^2/[(a+b)/2],0)
The center of the inverted circle is
      k^2     k^2
     ----- + -----
       a       b
    ---------------
           2
Therefore, for the center of the the inverted circle
to be the same as the image of the center of the given
circle we must have
      k^2     k^2
     ----- + -----
       a       b         k^2
    --------------- = ---------
           2            a + b
                       -------
                          2
               or
    (a - b)^2 = 0
Hence, the diameter of the given circle is zero.
See www.geocities.com/bractals/c-inv.jpg
 

Edited on February 25, 2006, 12:58 pm
  Posted by Bractals on 2006-02-25 12:08:38

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