% Inversion in the Plane
% Daniel Allcock

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\begin{document}

\title{Inversion in Circles}
\author{Daniel Allcock}
\address{Department of Mathematics\\University of Texas, Austin}
\email{daniel@allcock.org}
\urladdr{www.math.utexas.edu/users/allcock}
\date{Feb 19, 2020}
\maketitle

\noindent
We prove
the classical
fact that inversions in circles in the plane send
circles to circles (with lines understood as circles
of infinite radius).  Our proof uses classical 
pictures, but is simpler than usual because a
scaling argument lets one reduce to the essential cases.

Inversion in the unit circle $U$ means the self-map 
of $\R^2\cup\{\infty\}$ given by
$\chi:(r,\theta)\mapsto(1/r,\theta)$ in polar coordinates;
often we abbreviate $\chi(a)$ to $a'$.
It fixes $U$ pointwise and swaps the center
with $\infty$.
Inversion about circles of any radius,
centered at any point of $\R^2$, are
defined similarly (formally: by conjugating~$\chi$ 
by translations and scalings).  
To prove things about inversions in general circles
it is usually enough to prove them for~$\chi$.

\begin{lemma}
    Conjugation by $\chi$ inverts every scaling map
    $\lambda$.
    In particular, $\chi=\lambda\circ\chi\circ\lambda$.
    \qed
\end{lemma}

\begin{lemma}
    If  $a,b\in\R^2$ and $a',b'$ are their images
    under inversion about some circle with 
    center~$c\neq a,b$,
    then  $\triangle cab$ and $\triangle cb'a'$ are similar.
\end{lemma}

\begin{proof}
    Translations and scalings send circles to
    circles (and centers to centers) and respect
    triangle similarity.  So, by
    conjugating by them it
    is enough to prove this for~$\chi$. Rearranging
    $r_ar_{a'}=1=r_br_{b'}$ gives 
    $r_{a'}/r_{b'}=r_b/r_a$.  The  triangles
    have equal angles at~$0$, so they are similar.
\end{proof}

\begin{theorem}
    Inversion in any circle permutes the 
    circles not
    passing through its center.
\end{theorem}

\begin{proof}
    As before,
    it is enough to prove the theorem for~$\chi$.
    Suppose $C$ is a circle not containing~$0$. 
    Consider the  points of $C$
    closest to and furthest from~$0$.
    First we suppose $\chi$ swaps them,
    so we call them $a,a'$.  We claim that $\chi$ 
    preserves~$C$.  Suppose $b\in C$ and 
    consider the attached picture.  
    The triangle similarities let us define
    $$
    B:=\angle 0ab=\angle0b'a'
    \qquad
    s:=\angle 0a'b=\angle0b'a
    $$
    (mnemonic: {\larger Big} and {\smaller small}).
    As an angle inscribed in a semicircle,
    $\angle aba'$ is a right angle.
    Considering angles at~$a$ shows
    $B+(\pi/2-s)=\pi$, ie $B-s=\pi/2$.  Examining
    angles at~$b'$ shows that  $\angle ab'a'$  is a right
    angle (the dotted
    right angle symbol).  So $b'\in C$.

    For the general case we use a
    scaling trick.   Choose a scaling map
    $\lambda$ so that the points
    of $\lambda(C)$ nearest to and furthest from~$0$ are
    exchanged by~$\chi$. By the special case,
    $\chi$ preserves
    $\lambda(C)$, so $\chi(C)=\lambda(\chi(\lambda(C)))=
    \lambda(\lambda(C))$. This is a circle that misses~$0$.
\end{proof}

\begin{theorem}
    Inversion in a circle exchanges the circles 
    meeting the center with the lines (taken
    to 
    include~$\infty$) missing the center.
\end{theorem}

\begin{proof}
    As before, it is enough to prove the theorem
    for~$\chi$, and we begin with
    a special case.
    Choose $p\in U$, define $L$ and $C$ as the line
    and circle tangent to~$U$ at~$p$, with $C$ also
    required to contain~$0$.  We claim that $\chi$
    exchanges $C$ and~$L$.
    Suppose $a\neq0$ lies in~$C$,
    and write $a'$ for where the ray it generates 
    meets~$L$.  We must verify that $r_ar_{a'}=1$.
    As an angle  inscribed in a semicircle, 
    $\angle0ap$ is a right angle.  Therefore 
    $\triangle 0ap$ and $\triangle 0pa'$ are similar
    triangles.  Considering lengths of edges incident at~$0$
    gives $r_a/1=1/r_{a'}$.

    For the general case we use the same scaling
    trick. Let 
    $X$ denote either a circle containing~$0$ or a line
    missing~$0$.   Let
    $\lambda$ be the scaling map with
    $\lambda(X)$ tangent to~$U$.
    The special case shows that
    $\chi(\lambda(C))$ is a line missing~$0$, or
    a circle containing~$0$, depending on what $X$ was.
    so
    $
    \chi(C)=\lambda(\chi(\lambda(C)))
    $
    is also one.
\end{proof}

\end{document}
