Inverse of Circle Not Through Inversion Center/Proof 1

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Theorem

Let $\CC$ be a circle in the plane on center $O$.

Let $T : X \to Y$ be an inversive transformation with $\CC$ as the inversion circle.

Then $O$ is the inversion center.

Let $K$ be an arbitrary circle distinct from $\CC$ and not through $O$.


Then the image under $T$ of all the points on $K$ lie on the same circle, distinct from both $K$ and $\CC$.


Proof

Let $AB$ be the diameter of $K$, drawn so that $OAB$ are collinear points.

Let $A'$ and $B'$ be the image of $A$ and $B$ under $T$.

Let $K'$ be the circle with diameter $A'B'$

Let $C$ be an otherwise arbitrary point, lying on $K$.

Inverse Proof 4a2.png


\(\ds OA \cdot OA'\) \(=\) \(\ds OB \cdot OB' = OC \cdot OC'\) Definition of $T$
\(\ds OA : OC\) \(=\) \(\ds OC' : OA'\) rearranging
\(\ds \angle BOC\) \(=\) \(\ds \angle BOC\) shared
\(\ds \triangle OA'C'\) \(\sim\) \(\ds \triangle OCA\) Triangles with One Equal Angle and Two Sides Proportional are Similar
\(\ds \leadsto \ \ \) \(\ds \angle OA'C'\) \(=\) \(\ds \angle OCA = \alpha\)


\(\ds OB : OC\) \(=\) \(\ds OC' : OB'\) rearranging
\(\ds \triangle OC'B'\) \(\sim\) \(\ds \triangle OBC\) Triangles with One Equal Angle and Two Sides Proportional are Similar
\(\ds \leadsto \ \ \) \(\ds \angle OC'B'\) \(=\) \(\ds \angle OBC = \beta\)
\(\ds \leadsto \ \ \) \(\ds \gamma\) \(=\) \(\ds \angle OCB\)
\(\ds \gamma\) \(=\) \(\ds \alpha + \angle ACB\)


\(\ds \angle ACB\) \(=\) \(\ds 90^{\circ}\) Thales' Theorem
\(\ds \gamma\) \(=\) \(\ds \alpha + \angle A'C'B'\) External Angle of Triangle equals Sum of other Internal Angles
\(\ds \alpha + \angle A'C'B'\) \(=\) \(\ds \alpha + \angle ACB\) Common Notion $1$
\(\ds \angle A'C'B'\) \(=\) \(\ds \angle ACB\) Common Notion $3$
\(\ds \leadsto \ \ \) \(\ds \angle A'C'B'\) \(=\) \(\ds 90^{\circ}\) Common Notion $1$

By converse to Thales' Theorem:

$C'$ is on circle with diameter $A'B'$

We note two other arrangements both covered by this proof.

  • $K$ outside and tangent to $\CC$
  • $K$ cuts $\CC$ but not through $O$

The third possibility is simply the inverse of an inversive transformation.

  • $K$ inside $\CC$ and not through $O$


But this follows, since $T$ is an involution

$\blacksquare$


Sources


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