The inverses of (extended) lines and circles

To make the statement of the following theorem as simple as possible,
we use the notions of the point at infinity and extended lines.

The Inversion Theorem

Let C be a circle with centre O.
For any object X, let X' denote the inverse of X with respect to C.
  1. If L is an extended line through O, then L' = L
  2. If L is an extended line not through O, then L' is a circle through O.
  3. If D is a circle through O, then D' is an extended line through O
  4. If D is a circle not through O, then D' is a circle not through O

Proof

1. Suppose that P is a point on L (other that Å or O).
Then, by clause (1) of the definition of inversion, P' lies on L.
Since inversion in C interchanges O and Å, the result follows.

2. Let P be the foot of the perpendicular from O to L.
Suppose that Q is a point on L (other that Å or P), so <QPO is right.
By the Equal Angles Theorem, <QPO = <OQ'P', so <OQ'P' is a right angle.
Then Q' lies on the semicircle on OP' as diameter. since Å maps to O, we are done.

3. As inversion has order 2, (3) is equivalent to (2).

4. Let M denote the line joining O to the centre of D.
Suppose this cuts D at Q and R, so that QR is a diameter of D.
Now let P be a point on D, other than Q or R.
As an angle in a semicircle <QPR is a rigth angle.
As Q and R are on D, Q' and R' will lie on D'.
By the Corollary to the Equal Angles Theorem,
<:QPR = <R'P'Q', so the latter is also a right angle.
Thus, P' lies on the circle on Q'R' as diameter.
It follows that D' is precisely this circle.

CabriJava illustrations of cases 2 and 4

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