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 $𝔊$ 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 segmentwith endpoints $p$ and $q$ is denoted by $[p,q]$.

Before proving the property that a circle in $𝔊$ has a unique center,let us review some definitions.

Let $o$ and $a$ be points in $𝔊$, a geometry in which the congruence axioms are defined.Let $𝒞(o,a)$ be the set of all points $p$ in $𝔊$ such that theclosed line segments are congruent: $[o,a]\cong [o,p]$.The set $𝒞(o,a)$ is called a circle. When $a=o$, then $𝒞(o,a)$ is said to be degenerate.
Let $𝒞$ be a circle in $𝔊$. A center of $𝒞$ is a point $o$ such that for every pair of points $p,q$ in $𝒞$, $[o,p]\cong [o,q]$.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$𝒞(o,a)$ if .

We collect some simple facts below.

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

• Let $𝒞$ be a circle. If $o$ is a center of $𝒞$ and $a$ is any pointin $𝒞$, then $𝒞=𝒞(o,a)$, again by definition.

• A circle is degenerate if and only if it is a singleton.

  If $p$ is in $𝒞(o,o)$, then $[o,p]\cong [o,o]$, so that $p=o$, and$𝒞(o,o)=\{o\}$. Conversely, if $𝒞(o,a)=\{b\}$, 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 $𝒞(o,a)$ 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 $𝒞(o,a)=𝒞(o,o)$.

• Suppose $𝒞$ is a non-degenerate circle. Then every line passing through a center $o$ of $𝒞$ is incident with at least two points $a,a^{\prime }$ in $𝒞$. Furthermore, $o$ is the midpoint of $[a,a^{\prime }]$.

  If on $L$ through $o$ lies only one point $a\in 𝒞$, let $a^{\prime }$ be the point on the opposite ray of $\overrightarrow{oa}$ such that $a^{\prime }\in 𝒞$. Then $a^{\prime }=a$, which means that $o=a=a^{\prime }$, implying that $𝒞$ 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 $𝒞=𝒞(o,a)$ be a circle in $𝔊$.Suppose $o^{\prime }$ is another center of $𝒞$ and $o\ne o^{\prime }$. Let $L$ be the line passingthrough $o$ and $o^{\prime }$. Consider the (open) ray $\rho =\overrightarrow{o{o}^{\prime }}$. By one of the congruenceaxioms, there is a unique point $b$ on $\rho$ such that $[o,a]\cong [o,b]$.So $b\in 𝒞(o,a)$.

• 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 $d\in 𝒞(o,a)$. Since $d,o,o^{\prime }$ all lie on $L$, one must be between the other two.

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

  • 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 centerof $𝒞(o,a)$, 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:

  1. $d:o:o^{\prime }$, implying .

  2. $d:o:b$.

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

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

    1. 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.

    2. If $b:d:o^{\prime }$, then $d$ would be on $\rho$. Since $d$ is on $-\rho$, we have anothercontradiction.

    3. If $d:b:o^{\prime }$, then . But $o^{\prime }$ is a center of $𝒞(o,a)$, yet another contradiction.

    Therefore, Subcase 2 is impossible also.

  This means that Case 2 is impossible.

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

Remarks.

• The assumption that $𝔊$ 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 $𝒞(o,a)$ consists of exactly two points $a$ and $a^{\prime }$ (see figure above). In addition, $𝒞(o,a)$ 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 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.