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 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 geometry 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 congruent
: [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 midpoint of two points p and q
if [p,m]≅[m,q] and m,p,q are collinear
.
We say that p is an interior point of
𝒞(o,a) if [o,p]<[o,a].
We collect some simple facts below.
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•
In the circle 𝒞(o,a), o is a center of 𝒞(o,a) (by definition).
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•
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.
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•
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).
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•
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′ in 𝒞. Furthermore, o is the midpoint of [a,a′].
If on L through o lies only one point a∈𝒞, let a′ be the point on the opposite ray of →oa such that a′∈𝒞. 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.
Proof.
Let 𝒞=𝒞(o,a) be a circle in 𝔊. Suppose o′ is another center of 𝒞 and o≠o′. 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 b∈𝒞(o,a).
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•
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 d∈𝒞(o,a). Since d,o,o′ all lie on L, one must be between the other two.
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–
Subcase 1. If o:d:o′, then [o,d]<[o,o′]=[o,b]≅[o,a], contradiction
.
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Subcase 2. If d:o′:o, then [o,a]≅[o,b]=[o,o′]<[o,d], contradiction again.
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–
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 assumption
that o′ is not o in the beginning.
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–
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•
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:
-
(a)
d:o:o′, implying [o,d]<[o′,d].
-
(b)
d:o:b.
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–
Subcase 1. o:o′:b. Then [o′,b]<[o,b]≅[o,d]<[o′,d], contradiction.
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–
Subcase 2. o:b:o′. Let us look at the betweenness relations among the points b,d,o′.
-
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.
-
ii.
If b:d:o′, then d would be on ρ. Since d is on -ρ, we have another contradiction.
-
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.
-
i.
This means that Case 2 is impossible.
-
(a)
Since both Case 1 and Case 2 are impossible, o′=o, and the proof is complete. ∎
Remarks.
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•
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]≅[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 addition
, 𝒞(o,a) has two centers: o and o′.
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•
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.
References
-
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)
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 |