# 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 $\mathfrak{G}$ 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 circle. When $a=o$, then $\mathscr{C}(o,a)$ is said to be degenerate.
Let $\mathscr{C}$ be a circle in $\mathfrak{G}$. A 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 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 $\mathscr{C}(o,a)$ if $[o,p]<[o,a]$.

We collect some simple facts below.

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

• 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.

• 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 $\mathscr{C}(o,o)=\{o\}$. Conversely, if $\mathscr{C}(o,a)=\{b\}$, 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 $\mathscr{C}(o,a)$ 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 $\mathscr{C}(o,a)=\mathscr{C}(o,o)$.

• 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 $a,a^{\prime}$ in $\mathscr{C}$. Furthermore, $o$ is the midpoint of $[a,a^{\prime}]$.

If on $L$ through $o$ lies only one point $a\in\mathscr{C}$, let $a^{\prime}$ be the point on the opposite ray of $\overrightarrow{oa}$ such that $a^{\prime}\in\mathscr{C}$. Then $a^{\prime}=a$, which means that $o=a=a^{\prime}$, implying that $\mathscr{C}$ 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 $\mathfrak{G}$ has a unique center.

###### Proof.

Let $\mathscr{C}=\mathscr{C}(o,a)$ be a circle in $\mathfrak{G}$. Suppose $o^{\prime}$ is another center of $\mathscr{C}$ and $o\neq o^{\prime}$. Let $L$ be the line passing through $o$ and $o^{\prime}$. Consider the (open) ray $\rho=\overrightarrow{oo^{\prime}}$. By one of the congruence axioms, there is a unique point $b$ on $\rho$ such that $[o,a]\cong[o,b]$. So $b\in\mathscr{C}(o,a)$.

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

• 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. (a)

$d:o:o^{\prime}$, implying $[o,d]<[o^{\prime},d]$.

2. (b)

$d:o:b$.

• Subcase 1. $o:o^{\prime}:b$. Then $[o^{\prime},b]<[o,b]\cong[o,d]<[o^{\prime},d]$, contradiction.

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

1. 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.

2. ii.

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

3. iii.

If $d:b:o^{\prime}$, then $[o^{\prime},b]<[o^{\prime},d]$. But $o^{\prime}$ is a center of $\mathscr{C}(o,a)$, 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.

## References

• 1 M. J. Greenberg, , 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)
 Title proof of uniqueness of center of a circle Canonical name ProofOfUniquenessOfCenterOfACircle Date of creation 2013-03-22 17:17:02 Last modified on 2013-03-22 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