proof of uniqueness of center of a circle
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 $\U0001d50a$ 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 $\U0001d50a$ has a unique center, let us review some definitions.
Let $o$ and $a$ be points in $\U0001d50a$, a geometry^{} in which the congruence axioms are defined.
Let $\mathcal{C}(o,a)$ be the set of all points $p$ in $\U0001d50a$ such that the
closed line segments are congruent^{}: $[o,a]\cong [o,p]$.
The set $\mathcal{C}(o,a)$ is called a circle. When $a=o$, then $\mathcal{C}(o,a)$ is said to be degenerate.
Let $\mathcal{C}$ be a circle in $\U0001d50a$. A center of $\mathcal{C}$ is a point $o$ such that for every pair of points $p,q$ in $\mathcal{C}$, $[o,p]\cong [o,q]$.
We say that $m$ is a 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 interior point of
$\mathcal{C}(o,a)$ if $$.
We collect some simple facts below.

•
In the circle $\mathcal{C}(o,a)$, $o$ is a center of $\mathcal{C}(o,a)$ (by definition).

•
Let $\mathcal{C}$ be a circle. If $o$ is a center of $\mathcal{C}$ and $a$ is any point in $\mathcal{C}$, then $\mathcal{C}=\mathcal{C}(o,a)$, again by definition.

•
A circle is degenerate if and only if it is a singleton.
If $p$ is in $\mathcal{C}(o,o)$, then $[o,p]\cong [o,o]$, so that $p=o$, and $\mathcal{C}(o,o)=\{o\}$. Conversely, if $\mathcal{C}(o,a)=\{b\}$, 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 $\mathcal{C}(o,a)$ 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 $\mathcal{C}(o,a)=\mathcal{C}(o,o)$.

•
Suppose $\mathcal{C}$ is a nondegenerate circle. Then every line passing through a center $o$ of $\mathcal{C}$ is incident^{} with at least two points $a,{a}^{\prime}$ in $\mathcal{C}$. Furthermore, $o$ is the midpoint of $[a,{a}^{\prime}]$.
If on $L$ through $o$ lies only one point $a\in \mathcal{C}$, let ${a}^{\prime}$ be the point on the opposite ray of $\overrightarrow{oa}$ such that ${a}^{\prime}\in \mathcal{C}$. Then ${a}^{\prime}=a$, which means that $o=a={a}^{\prime}$, implying that $\mathcal{C}$ is degenerate. Since $[o,a]\cong [o,{a}^{\prime}]$, and $o,a,{a}^{\prime}$ lie on the same line, $o$ is the midpoint of $[a,{a}^{\prime}]$.
Now, on to the main fact.
Theorem 1.
Every circle in $\mathrm{G}$ has a unique center.
Proof.
Let $\mathcal{C}=\mathcal{C}(o,a)$ be a circle in $\U0001d50a$. Suppose ${o}^{\prime}$ is another center of $\mathcal{C}$ and $o\ne {o}^{\prime}$. Let $L$ be the line passing through $o$ and ${o}^{\prime}$. Consider the (open) ray $\rho =\overrightarrow{o{o}^{\prime}}$. By one of the congruence axioms, there is a unique point $b$ on $\rho $ such that $[o,a]\cong [o,b]$. So $b\in \mathcal{C}(o,a)$.

•
Case 1. Suppose ${o}^{\prime}=b$. Consider the (open) opposite ray $\rho $ of $\rho $. There is a unique point $d$ on $\rho $ such that $[o,d]\cong [o,a]$. So $d\in \mathcal{C}(o,a)$. Since $d,o,{o}^{\prime}$ all lie on $L$, one must be between the other two.

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Subcase 1. If $o:d:{o}^{\prime}$, then $$, contradiction^{}.

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Subcase 2. If $d:{o}^{\prime}:o$, then $$, contradiction again.

–
Subcase 3. So suppose $o$ is between $d$ and ${o}^{\prime}$. Now, since ${o}^{\prime}$ is also a center of $\mathcal{C}(o,a)$, we have that $[b,b]=[{o}^{\prime},b]\cong [{o}^{\prime},d]$, which implies that ${o}^{\prime}=d$ by another one of the congruence axioms. But $d:o:{o}^{\prime}$, which forces ${o}^{\prime}=o$, contradicting the assumption^{} that ${o}^{\prime}$ is not $o$ in the beginning.

–

•
Case 2. If ${o}^{\prime}$ is not $b$, then since $o,{o}^{\prime},b$ 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 ${o}^{\prime}$ and $b$. So we have only two subcases to deal with: either $o:{o}^{\prime}:b$, or $o:b:{o}^{\prime}$. 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:

(a)
$d:o:{o}^{\prime}$, implying $$.

(b)
$d:o:b$.

–
Subcase 1. $o:{o}^{\prime}:b$. Then $$, contradiction.

–
Subcase 2. $o:b:{o}^{\prime}$. Let us look at the betweenness relations among the points $b,d,{o}^{\prime}$.

i.
If $b:{o}^{\prime}:d$, then $d:{o}^{\prime}:o$ by one of the conditions of the betweenness relations. But this forces ${o}^{\prime}$ to be on $\rho $. Since ${o}^{\prime}$ is on $\rho $, this is a contradiction.

ii.
If $b:d:{o}^{\prime}$, then $d$ would be on $\rho $. Since $d$ is on $\rho $, we have another contradiction.

iii.
If $d:b:{o}^{\prime}$, then $$. But ${o}^{\prime}$ is a center of $\mathcal{C}(o,a)$, yet another contradiction.
Therefore, Subcase 2 is impossible also.

i.
This means that Case 2 is impossible.

(a)
Since both Case 1 and Case 2 are impossible, ${o}^{\prime}=o$, and the proof is complete^{}. ∎
Remarks.

•
The assumption that $\U0001d50a$ is ordered cannot be dropped. Here is a simple counterexample. Consider an incidence geometry defined on a circle $C$ in the Euclidean plane^{}.
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 $\mathcal{C}(o,a)$ consists of exactly two points $a$ and ${a}^{\prime}$ (see figure above). In addition^{}, $\mathcal{C}(o,a)$ has two centers: $o$ and ${o}^{\prime}$.

•
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 nonArchimedean, 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.
References
 1 M. J. Greenberg, Euclidean^{} and NonEuclidean Geometries, Development and History, W. H. Freeman and Company, San Francisco (1974)
 2 N. Koblitz, padic Numbers, padic Analysis, and ZetaFunctions, SpringerVerlag, New York (1977)
Title  proof of uniqueness of center of a circle 
Canonical name  ProofOfUniquenessOfCenterOfACircle 
Date of creation  20130322 17:17:02 
Last modified on  20130322 17:17:02 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  32 
Author  CWoo (3771) 
Entry type  Proof 
Classification  msc 51M10 
Classification  msc 51M04 
Classification  msc 51G05 
Related topic  Midpoint4 
Defines  midpoint 
Defines  circle 
Defines  interior point 