Inverse of Circle Not Through Inversion Center/Proof 2
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 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
- 1996: Richard Courant, Herbert Robbins and Ian Stewart: What is Mathematics? (2nd ed.): Chapter $\text{III}$ / $\text{II}$ Section $4$: "Geometrical Transformations. Inversion."