Inverse of a Line Through Circle Center is a Line

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Theorem

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

Let the radius of $C$ be $r$.

Let $L$ be an arbitrary ray through $O$.

Let $P$ be an otherwise arbitrary point, but lying on $L$.

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

Then $O$ is the inversion center.

Then the image of $P$ under $T$ also lies on $L$.

Proof

Let $P'$ be the image of $P$ under $T$, with $P \ne O$.

By the definition of inversion:

$P'$ lies on $L$

We deal with $O$ separately. By the definition of ray:

$O$ lies on $L$

In the words of Euclid:

A point is that which has no part.

A point has no length, it has no magnitude.

So we cannot compute $\dfrac {r^2} {OO}$ to find $O'$, the image of $O$.

But by the definition of inversion, the image of $O$ is the point at infinity.

By the definition the point at infinity lies on every infinite half-line through $O$.

Hence $O$ lies on $L$.

Therefore the image under $T$ of every point on $L$ lies on $L$.

$\blacksquare$