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 geometryMathworldPlanetmath satisfying the congruence axiomsMathworldPlanetmath. 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 𝔊 has a unique center, let us review some definitions.

Let o and a be points in 𝔊, a geometryMathworldPlanetmath in which the congruence axioms are defined. Let 𝒞(o,a) be the set of all points p in 𝔊 such that the closed line segments are congruentMathworldPlanetmathPlanetmath: [o,a][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][o,q]. We say that m is a midpointMathworldPlanetmathPlanetmathPlanetmath of two points p and q if [p,m][m,q] and m,p,q are collinearMathworldPlanetmath.
We say that p is an interior point of 𝒞(o,a) if [o,p]<[o,a].

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 point in 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][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 ρ on L emanating from o. Then there is a point d on ρ such that [o,d][o,a]. So d=a since 𝒞(o,a) is a singleton containing a. Similarly, there is a unique e on -ρ, the opposite ray of ρ, with [o,e][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 incidentMathworldPlanetmath with at least two points a,a in 𝒞. Furthermore, o is the midpoint of [a,a].

    If on L through o lies only one point a𝒞, let a be the point on the opposite ray of oa such that a𝒞. Then a=a, which means that o=a=a, implying that 𝒞 is degenerate. Since [o,a][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 G has a unique center.


Let 𝒞=𝒞(o,a) be a circle in 𝔊. Suppose o is another center of 𝒞 and oo. Let L be the line passing through o and o. Consider the (open) ray ρ=oo. By one of the congruence axioms, there is a unique point b on ρ such that [o,a][o,b]. So b𝒞(o,a).

  • Case 1. Suppose o=b. Consider the (open) opposite ray -ρ of ρ. There is a unique point d on -ρ such that [o,d][o,a]. So d𝒞(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][o,a], contradictionMathworldPlanetmathPlanetmath.

    • Subcase 2. If d:o:o, then [o,a][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 𝒞(o,a), we have that [b,b]=[o,b][o,d], which implies that o=d by another one of the congruence axioms. But d:o:o, which forces o=o, contradicting the assumptionPlanetmathPlanetmath 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 ρ 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 -ρ of ρ with d on -ρ such that [o,d][o,a][o,b]. From the properties of opposite rays, we also have the following two facts:

    1. (a)

      d:o:o, implying [o,d]<[o,d].

    2. (b)


    • Subcase 1. o:o:b. Then [o,b]<[o,b][o,d]<[o,d], contradiction.

    • Subcase 2. o:b:o. Let us look at the betweenness relations among the points b,d,o.

      1. i.

        If b:o:d, then d:o:o by one of the conditions of the betweenness relations. But this forces o to be on -ρ. Since o is on ρ, this is a contradiction.

      2. ii.

        If b:d:o, then d would be on ρ. Since d is on -ρ, we have another contradiction.

      3. iii.

        If d:b:o, then [o,b]<[o,d]. But o 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=o, and the proof is completePlanetmathPlanetmathPlanetmathPlanetmath. ∎


  • 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 planeMathworldPlanetmath.

    It is not possible to define a betweenness relation on C. However, it is still possible to define a congruence relationPlanetmathPlanetmath on C: [x,y][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 (see figure above). In additionPlanetmathPlanetmath, 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 triangleMathworldPlanetmath 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.


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