proof of uniqueness of center of a circleProofOfUniquenessOfCenterOfACircle2007-06-19 17:37:192007-10-25 16:05:36Proofcircle has one center0midpointcircleinterior point\usepackage{amssymb,amscd}
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In this entry, we prove the uniqueness of center of a circle in a slightly more general setting than the parent entry.
In this more general setting, let $\mathfrak{G}$ be an ordered geometry satisfying the congruence axioms.
We write $a:b:c$ to mean $b$ is between $a$ and $c$. Recall that the closed line segment
with endpoints $p$ and $q$ is denoted by $[p,q]$.
Before proving the property that a circle in $\mathfrak{G}$ has a unique center,
let us review some definitions.
Let $o$ and $a$ be points in $\mathfrak{G}$, a geometry in which the congruence axioms are defined.
Let $\mathscr{C}(o,a)$ be the set of all points $p$ in $\mathfrak{G}$ such that the
closed line segments are congruent: $[o,a]\cong [o,p]$.
The set $\mathscr{C}(o,a)$ is called a \emph{circle}. When $a=o$, then $\mathscr{C}(o,a)$ is said to be \emph{degenerate}. \\
Let $\mathscr{C}$ be a circle in $\mathfrak{G}$. A \emph{center} of $\mathscr{C}$ is a point $o$ such that for every pair of points $p,q$ in $\mathscr{C}$, $[o,p] \cong [o,q]$.
We say that $m$ is a \emph{midpoint} of two points $p$ and $q$
if $[p,m] \cong [m,q]$ and $m,p,q$ are collinear. \\
We say that $p$ is an \emph{interior point} of
$\mathscr{C}(o,a)$ if $[o,p] < [o,a]$.
We collect some simple facts below.
\begin{itemize}
\item In the circle $\mathscr{C}(o,a)$, $o$ is a center of $\mathscr{C}(o,a)$ (by definition).
\item Let $\mathscr{C}$ be a circle. If $o$ is a center of $\mathscr{C}$ and $a$ is any point
in $\mathscr{C}$, then $\mathscr{C}=\mathscr{C}(o,a)$, again by definition.
\item A circle is degenerate if and only if it is a singleton.
If $p$ is in $\mathscr{C}(o,o)$, then $[o,p]\cong [o,o]$, so that $p=o$, and
$\mathscr{C}(o,o)=\lbrace o\rbrace$. Conversely, if $\mathscr{C}(o,a)=\lbrace b\rbrace$, then $b=a$. Let $L$ be any line passing through $o$. Choose a ray $\rho$ on $L$ emanating from $o$. Then there is a point $d$ on $\rho$ such that $[o,d]\cong [o,a]$. So $d=a$ since $\mathscr{C}(o,a)$ is a singleton containing $a$.
Similarly, there is a unique $e$ on $-\rho$, the opposite ray of $\rho$, with $[o,e]\cong [o,a]$.
So $e=a$. Since $d:o:e$, we have that $a=d=o$. Therefore $\mathscr{C}(o,a)=\mathscr{C}(o,o)$.
\item Suppose $\mathscr{C}$ is a non-degenerate circle. Then every line passing through a center $o$ of $\mathscr{C}$ is incident with at least two points $a,a'$ in $\mathscr{C}$. Furthermore, $o$ is the midpoint of $[a,a']$.
If on $L$ through $o$ lies only one point $a \in \mathscr{C}$, let $a'$ be the point on the opposite ray of $\ray{oa}$ such that $a'\in \mathscr{C}$. Then $a'=a$, which means that $o=a=a'$, implying that $\mathscr{C}$ is degenerate. Since $[o,a]\cong [o,a']$, and $o,a,a'$ lie on the same line, $o$ is the midpoint of $[a,a']$.
\end{itemize}
Now, on to the main fact.
\begin{thm} Every circle in $\mathfrak{G}$ has a unique center.
\end{thm}
\begin{proof} Let $\mathscr{C}=\mathscr{C}(o,a)$ be a circle in $\mathfrak{G}$.
Suppose $o'$ is another center of $\mathscr{C}$ and $o\ne o'$. Let $L$ be the line passing
through $o$ and $o'$. Consider the (open) ray $\rho = \ray{oo'}$. By one of the congruence
axioms, there is a unique point $b$ on $\rho$ such that $[o,a] \cong [o,b]$.
So $b\in \mathscr{C}(o,a)$.
\begin{itemize}
\item Case 1. Suppose $o'=b$. Consider the (open) opposite ray $-\rho$ of $\rho$.
There is a unique point $d$ on $-\rho$ such that $[o,d]\cong [o,a]$. So $d\in\mathscr{C}(o,a)$. Since $d,o,o'$ all lie on $L$, one must be between the other two.
\begin{itemize}
\item
Subcase 1. If $o:d:o'$, then $[o,d] < [o,o'] = [o,b] \cong [o,a]$, contradiction.
\item
Subcase 2. If $d:o':o$, then $[o,a] \cong [o,b] = [o,o'] < [o,d]$, contradiction again.
\item
Subcase 3. So suppose $o$ is between $d$ and $o'$. Now, since $o'$ is also a center
of $\mathscr{C}(o,a)$, we have that $[b,b] = [o',b] \cong [o',d]$, which implies that
$o'=d$ by another one of the congruence axioms. But $d:o:o'$, which forces $o'=o$, contradicting the assumption that $o'$ is not $o$ in the beginning.
\end{itemize}
\item
Case 2. If $o'$ is not $b$, then since $o,o',b$ lie on the same line $L$, one must be between the other two. Since $b$ also lies on the ray $\rho$ with $o$ as the source, $o$ cannot be between $o'$ and $b$. So we have only two subcases to deal with: either $o:o':b$, or $o:b:o'$. In either subcase, we need to again consider the opposite ray $-\rho$ of $\rho$ with $d$ on $-\rho$ such that $[o,d]\cong [o,a] \cong [o,b]$.
From the properties of opposite rays, we also have the following two facts:
\begin{enumerate}
\item $d:o:o'$, implying $[o,d] < [o',d]$.
\item $d:o:b$.
\end{enumerate}
\begin{itemize}
\item
Subcase 1. $o:o':b$. Then $[o',b] < [o,b] \cong [o,d] < [o',d]$, contradiction.
\item
Subcase 2. $o:b:o'$. Let us look at the betweenness relations among the points $b, d, o'$.
\begin{enumerate}
\item If $b:o':d$, then $d:o':o$ by one of the conditions of the betweenness relations. But this forces $o'$ to be on $-\rho$. Since $o'$ is on $\rho$, this is a contradiction.
\item If $b:d:o'$, then $d$ would be on $\rho$. Since $d$ is on $-\rho$, we have another
contradiction.
\item If $d:b:o'$, then $[o',b] < [o',d]$. But $o'$ is a center of $\mathscr{C}(o,a)$, yet another contradiction.
\end{enumerate}
Therefore, Subcase 2 is impossible also.
\end{itemize}
This means that Case 2 is impossible.
\end{itemize}
Since both Case 1 and Case 2 are impossible, $o'=o$, and the proof is complete. \end{proof}
\textbf{Remarks}.
\begin{itemize}
\item
The assumption that $\mathfrak{G}$ is ordered cannot be dropped. Here is a simple counterexample. Consider an incidence geometry defined on a circle $C$ in the Euclidean plane.
\begin{center}
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It is not possible to define a betweenness relation on $C$. However, it is still possible to define a congruence relation on $C$: $[x,y]\cong [z,t]$ if $[x,y]$ and $[z,t]$ have the same arc length. Given any points $o,a$ on $C$, the circle $\mathscr{C}(o,a)$ consists of exactly two points $a$ and $a'$ (see figure above). In addition, $\mathscr{C}(o,a)$ has two centers: $o$ and $o'$.
\item
There is another definition of a circle, which is based on the concept of a metric. In this definition, examples can also be found where the uniqueness of center of a circle fails. The most commonly quoted example is found in the metric space of $p$-adic numbers. The metric defined is non-Archimedean, so every triangle is isosceles (see the note in ultrametric triangle inequality). From this it is not hard to see that every interior point of a circle is its center.
\end{itemize}
\begin{thebibliography}{6}
\bibitem{mg} M. J. Greenberg, {\it Euclidean and Non-Euclidean Geometries, Development and History}, W. H. Freeman and Company, San Francisco (1974)
\bibitem{nk} N. Koblitz, {\it p-adic Numbers, p-adic Analysis, and Zeta-Functions}, Springer-Verlag, New York (1977)
\end{thebibliography}