Inverse of Circle Through Inversion Center is Straight Line Not Through Inversion Center

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Theorem

Let an arbitrary circle $K$ be drawn in the plane.

Let $A'$ and $P'$ be arbitrary points on $K$.

Let $T$ be an inversive transformation such that:

the inversion center of $T$ is $O$
the inversion circle $O$ for $T$ is chosen such that $OA'$ is a diameter of $K$
the radius of $O$ is $r$.

Let $A$ and $P$ be the images of $A'$ and $P'$ under $T$ respectively.

Let $L$ be a straight line drawn perpendicular to the extension of $OA'$ at point $A$.

Then $P$ lies on $L$.


Proof

The diagrams show the two cases:

$K$ completely inside the circle on $O$


CircleInverse1.png


or with some points outside


CircleInverse2.png


The proof is the same for both cases.

\(\ds OP \cdot OP'\) \(=\) \(\ds OA \cdot OA'\) definition of $T$
\(\ds \leadsto \ \ \) \(\ds \dfrac {OP} {OA}\) \(=\) \(\ds \dfrac {OA'} {OP'}\) rearranging

$\angle AOP$ is shared.

\(\ds \triangle OAP\) \(\sim\) \(\ds \triangle OP'A'\) Triangles with One Equal Angle and Two Sides Proportional are Similar
\(\ds \leadsto \ \ \) \(\ds \angle OAP\) \(=\) \(\ds 90 \degrees\) Definition of Perpendicular

By Perpendicular from Point to Straight Line in Plane is Unique:

there is only one straight line perpendicular to a straight line at a given point.


So:

$AP$ is the same straight line as $L$

That is, $P$ lies on $L$.

$P$ is the image under $T$ of an arbitrary point $P'$ on $K$, and $P$ lies on $L$.

$\blacksquare$


Sources

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