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 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 is called a circle. When $a=o$ , then is said to be degenerate.
Let be a circle in $\mathfrak{G}$ . A center of is a point $o$ such that for every pair of points $p,q$ in , $[o,p] \cong [o,q]$ . 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 if $[o,p] < [o,a]$ .

We collect some simple facts below.

• In the circle , $o$ is a center of (by definition).
• Let be a circle. If $o$ is a center of and $a$ is any point in , then , again by definition.
• A circle is degenerate if and only if it is a singleton.

  If $p$ is in , then $[o,p]\cong [o,o]$ , so that $p=o$ , and . Conversely, if , 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 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 .

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

  If on $L$ through $o$ lies only one point , let $a'$ be the point on the opposite ray of $\ray{oa}$ such that . Then $a'=a$ , which means that $o=a=a'$ , implying that is degenerate. Since $[o,a]\cong [o,a']$ , and $o,a,a'$ lie on the same line, $o$ is the midpoint of $[a,a']$ .

Now, on to the main fact.

Proof. Let be a circle in $\mathfrak{G}$ . Suppose $o'$ is another center of 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 .

• 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 . Since $d,o,o'$ all lie on $L$ , one must be between the other two.
  • Subcase 1. If $o:d:o'$ , then $[o,d] < [o,o'] = [o,b] \cong [o,a]$ , contradiction.
  • Subcase 2. If $d:o':o$ , then $[o,a] \cong [o,b] = [o,o'] < [o,d]$ , contradiction again.
  • Subcase 3. So suppose $o$ is between $d$ and $o'$ . Now, since $o'$ is also a center of , 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.
• 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:
  1. $d:o:o'$ , implying $[o,d] < [o',d]$ .
  2. $d:o:b$ .
  • Subcase 1. $o:o':b$ . Then $[o',b] < [o,b] \cong [o,d] < [o',d]$ , contradiction.
  • Subcase 2. $o:b:o'$ . Let us look at the betweenness relations among the points $b, d, o'$ .
    1. 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.
    2. If $b:d:o'$ , then $d$ would be on $\rho$ . Since $d$ is on $-\rho$ , we have another contradiction.
    3. If $d:b:o'$ , then $[o',b] < [o',d]$ . But $o'$ is a center of , 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, $o'=o$ , and the proof is complete.

Remarks.

• 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.
  
  
  

  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 consists of exactly two points $a$ and $a'$ (see figure above). In addition, has two centers: $o$ and $o'$ .

• 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.

Bibliography

1

M. J. Greenberg, Euclidean and Non-Euclidean Geometries, Development and History, W. H. Freeman and Company, San Francisco (1974)

2

N. Koblitz, p-adic Numbers, p-adic Analysis, and Zeta-Functions, Springer-Verlag, New York (1977)