Theorem: (Line Inversion)

Let K be a circle with center S. Let C be a circle passing through

S and SL₁ be the diameter of C passing through S. Let L₂ be the

inverse of L₁ w.r.t. circle K. Then the inverse of C w.r.t. circle

K is the line through L₂ perpendicular to SL₁. Conversely, given

a straight line, l, which does not pass through S, let L₂ be the point

where the perpendicular from S meets l. Then the inverse of l w.r.t.

K is a circle which contains S and whose diameter through S

terminates at point L₁, where L₁ is the inverse of L₂.

Proof:

Part 1

We want to show that the inverse of C is contained in l.

Let j be the line containing S, L₁, and L₂.

Construct point T', where K and j intersect, so that ST' is a radius, r, of K.

Construct P₁ on C, where P₁ does not equal L₁. Also let P₂ be the inverse

of P₁.

Construct ray SP₂ containing P₁.

Construct point T, where K and SP₂ intersect, so that ST = r.

By the definition of Inverse Image:

(ST')² = SL₁ * SL₂ and

(ST)² = SP₁ * SP₂.

Since ST' = r = ST, we get

SL₁ * SL₂ = SP₁ * SP₂.

Which gives the following relationship

SL₁/SP₁ = SP₂/SL₂.

Since the sides, SL₁, SL₂, SP₁, and SP₂, of the triangles SP₂L₂ and SL₁P₁ are

proportional and share a comman angle at the vertex S they are similar by

S.A.S. for similar triangles.

We know that angle SP₁L₁ is a right angle because it is inscribed in a

semi-circle.

Since angle SP₁L₁ corresponds to angle SL₂P₂, angle SL₂P₂ is also a

right angle.

Hence l is perpendicular to j.

Therefore P₂ lies on l.

The following defines the notation being used.

Defintion: Let I be the inversion map:

I(P) = P', where P' is the inversion of point P and

similarly for any set S, let I(S) = S' where

S' = {P' | P' = I(P) for some P belonging to S}.

Hence we have shown that

I(P₁) = P₂, which lies on l.

Therefore I(C) is contained in l.

Part 2

We want to show that the inverse of C contains l.

Now pick any point Q on l, as seen in the following diagram.



Construct ray SQ containing point Q₁ not equal to S on C.

If Q₁ is equal to S then SQ is tangent to C at S.

Therefore SQ would be parallel to l,

which is a contradiction since Q is on l and Q₁ is on SQ and C.

Hence by part 1,

I(Q₁) = Q,

so I(C) contains l.

Therefore I(C) = l.

Coversely

We want to show that

I(l) = C.

Since we know that

I(C) = l,

if we take the inversion of I(C) = l we get

I(I(C)) = I(l)

C = I(l).

Q.E.D.