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[parent] proof of uniqueness of center of a circle (Proof)

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)=\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)$.

  • 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 $ \overrightarrow{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']$.

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'$ 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 = \overrightarrow{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)$.
  • 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.
    • 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 $ \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.
  • 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 $ \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'=o$, and the proof is complete. $ \qedsymbol$

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.

    \begin{pspicture} % latex2html id marker 182 (-4,-2.5)(4,2.5) \rput[a](0,2){.} \... ...ut[l](2.2,0){$o'$} \rput[r](-1.9,1){$a$} \rput[r](-1.9,-1){$a'$} \end{pspicture}

    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'$.

  • 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)



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Also defines:  midpoint, circle, interior point

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Cross-references: ultrametric triangle inequality, isosceles, triangle, non-archimedean, metric space, metric, addition, arc length, congruence relation, Euclidean plane, counterexample, betweenness relations, source, implies, contradiction, open, lie on, incident, non-degenerate, opposite ray, ray, passing through, line, singleton, collinear, congruent, geometry, points, definitions, property, endpoints, closed line segment, mean, congruence axioms, ordered geometry, parent, center
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This is version 29 of proof of uniqueness of center of a circle, born on 2007-06-19, modified 2007-10-25.
Object id is 9625, canonical name is ProofOfUniquenessOfCenterOfACircle.
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Classification:
AMS MSC: 51G05 (Geometry :: Ordered geometries )
 51M04 (Geometry :: Real and complex geometry :: Elementary problems in Euclidean geometries)
 51M10 (Geometry :: Real and complex geometry :: Hyperbolic and elliptic geometries and generalizations)
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