(b) For any points X, Y ,
8.5 TRANSFORMATIONS 309
X'y' =
Xy
r^2.
OX·Oy
(c) The image of any "circle" is a "circle."
(d) The image of any (ordinary) circle y that does not intersect 0 is an image of y
by a homothety centered at O.
Statements (a) and (b) are easy to prove, using similar triangles, and (d) is a simple
consequence of (c). Even (c), remarkable as it is, is not too hard to prove. There are
several cases to consider. Let us consider one.
Example 8.5.15 Let co be a circle with center 0, and let y be a circle that passes
through 0 and intersects co in points A and B. Under inversion with respect to co, the
image of y is the line passing through A and B.
Solution: Let X be a point on yand outside co, as shown below. Using the algo
rithm on page 307, we draw the tangents from X to co, intersecting CO at C and D; the
midpoint of CD is the image X'. We wish to show that X' lies on AB.
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This is easy, since we are hands-on geometers. How did we draw those tangents in the
first place? We erected a right triangle with hypotenuse on OX and right angle C on co.
To do this, we drew a circle with diameter OX; it intersected co at C and D. In other
words, implicit in our diagram are three circles: co, y, and the circle 1C with diameter
OX.
Now CD is a common chord of CO and 1C; OX is a common chord of 1C and r. and
AB is common to CO and y. That is exactly the situation of Example 8.4.6 on page 292;
so we conclude that these three chords are concurrent. Hence X' must lie on AB. •