proof of uniqueness of center of a circle

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.

•

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

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

o

⁢

a

→

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

=

o

⁢

o

′

→

. 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

)

.

•

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.

–

Subcase 1. If

o

:

d

:

o

′

, then

[

o

,

d

]

<

[

o

,

o

′

]

=

[

o

,

b

]

≅

[

o

,

a

]

,

contradiction

.

–

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

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

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

.

–

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

′

.

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.

This means that Case 2 is impossible.

Since both Case 1 and Case 2 are impossible,

o

′

=

o

, and the proof is

complete

. ∎

Remarks

.

•

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

′

.

•

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

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