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.
Before proving the property that a circle in has a unique center, let us review some definitions.
Let and be points in , a geometry in which the congruence axioms are defined.
Let be the set of all points in such that the
closed line segments are congruent: .
The set is called a circle. When , then is said to be degenerate.
Let be a circle in . A center of is a point such that for every pair of points in , . We say that is a midpoint of two points and if and are collinear.
We say that is an interior point of if .
We collect some simple facts below.
In the circle , is a center of (by definition).
Let be a circle. If is a center of and is any point in , then , again by definition.
A circle is degenerate if and only if it is a singleton.
If is in , then , so that , and . Conversely, if , then . Let be any line passing through . Choose a ray on emanating from . Then there is a point on such that . So since is a singleton containing . Similarly, there is a unique on , the opposite ray of , with . So . Since , we have that . Therefore .
If on through lies only one point , let be the point on the opposite ray of such that . Then , which means that , implying that is degenerate. Since , and lie on the same line, is the midpoint of .
Now, on to the main fact.
Every circle in has a unique center.
Let be a circle in . Suppose is another center of and . Let be the line passing through and . Consider the (open) ray . By one of the congruence axioms, there is a unique point on such that . So .
Case 1. Suppose . Consider the (open) opposite ray of . There is a unique point on such that . So . Since all lie on , one must be between the other two.
Subcase 1. If , then , contradiction.
Subcase 2. If , then , contradiction again.
Subcase 3. So suppose is between and . Now, since is also a center of , we have that , which implies that by another one of the congruence axioms. But , which forces , contradicting the assumption that is not in the beginning.
Case 2. If is not , then since lie on the same line , one must be between the other two. Since also lies on the ray with as the source, cannot be between and . So we have only two subcases to deal with: either , or . In either subcase, we need to again consider the opposite ray of with on such that . From the properties of opposite rays, we also have the following two facts:
, implying .
Subcase 1. . Then , contradiction.
Subcase 2. . Let us look at the betweenness relations among the points .
If , then by one of the conditions of the betweenness relations. But this forces to be on . Since is on , this is a contradiction.
If , then would be on . Since is on , we have another contradiction.
If , then . But 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, , and the proof is complete. ∎
The assumption that is ordered cannot be dropped. Here is a simple counterexample. Consider an incidence geometry defined on a circle in the Euclidean plane.
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 -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.
|Title||proof of uniqueness of center of a circle|
|Date of creation||2013-03-22 17:17:02|
|Last modified on||2013-03-22 17:17:02|
|Last modified by||CWoo (3771)|