Inverse of Circle Not Through Inversion Center

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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 1

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

$\blacksquare$


Proof 2

Inverse Proof 4b.png

Let the radius of $K$ be $k$.

Draw an arbitrary straight line from $O$ cutting $K$ at $A$ and $B$.

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

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

There are two cases:

$(1): \quad$ The inversion circle $C$ may pass through straight line $OAB$ outside $K$
$(2): \quad$ $C$ may cut $K$ and pass inside it.

The second possibility is illustrated using a dotted line.

The proof is the same for both.

Note: the inversion circle is not shown, for clarity.

Let:

$OA = a$
$OB = b$
$OA' = a'$
$OB' = b'$
$OM = m$

Let $t$ be the length of the tangent line to $K$ from $O$.

\(\ds a a'\) \(=\) \(\ds b b' = r^2\) Definition of $T$
\(\ds ab\) \(=\) \(\ds t^2\) Tangent Secant Theorem
\(\ds \dfrac {a'} b\) \(=\) \(\ds \dfrac {b'} a = \dfrac {r^2} {t^2}\) dividing by $t^2$

Let the constant:

$\dfrac {r^2} {t^2} = c^2$


Draw $A'Q \parallel BM$.

Let $OQ = q$.

Let $A'Q = p$.

By Equiangular Triangles are Similar:

$\triangle OQA' \sim \triangle OMB$
$\leadsto \dfrac q m = \dfrac {a'} b = c^2$

Also

$\leadsto \dfrac p k = \dfrac {a'} b = c^2$

Substituting:

$q = m c^2$

and

$p = k c^2$

Since the right hand side of both of these are constant, for all positions of $A$ and $B$:

$Q$ is the same point on $OM$

The length $A'Q = p$ is constant.

$A'$ is always the same distance from $Q$.

Recall:

$\dfrac {b'} a = c^2$

So mutatis mutandis:

$B'Q = p$

$B'$ is also always the same distance from $Q$.

So $B'Q = A'Q$.

It follows that $A'$ and $B'$ both lie on the same circle.

$\blacksquare$


Sources

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