# Finding Center of Circle

## Contents

## Theorem

For any given circle, it is possible to find its center.

In the words of Euclid:

(*The Elements*: Book $\text{III}$: Proposition $1$)

## Proof 1

Draw any chord $AB$ on the circle in question.

Bisect $AB$ at $D$.

Construct $CE$ perpendicular to $AB$ at $D$, where $C$ and $E$ are where this perpendicular meets the circle.

Bisect $CE$ at $F$.

Then $F$ is the center of the circle.

The proof is as follows.

Suppose $F$ were not the center of the circle, but that $G$ were instead.

Join $GA, GB, GD$.

As $G$ is (as we have supposed) the center, then $GA = GB$.

Also, we have $DA = DB$ as $D$ bisects $AB$.

So from Triangle Side-Side-Side Equality:

- $\triangle ADG = \triangle BDG$

Hence:

- $\angle ADG = \angle BDG$

But from Book $\text{I}$ Definition $10$: Right Angle:

*When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is***right**, and the straight line standing on the other is called a**perpendicular**to that on which it stands.

So $\angle ADG$ is a right angle.

But $\angle ADF$ is also a right angle.

So $\angle ADG = \angle ADF$, and this can happen only if $G$ lies on $CE$.

But then as $G$ is, as we suppose, at the center of the circle, then $GC = GE$.

Thus it follows that $G$ bisects $CE$.

But then $GC = FC$, and so $G = F$.

Hence the result.

$\blacksquare$

### Porism

In the words of Euclid:

*From this it is manifest that, if in a circle a straight line cut a straight line into two equal parts and at right angles, the centre of the circle is on the cutting straight line.*

(*The Elements*: Book $\text{III}$: Proposition $1$ : Porism)

## Proof 2

From Perpendicular Bisector of Chord Passes Through Center, $CE$ passes through the center of the circle.

The center must be the point $F$ such that $FE = FC$.

That is, $F$ is the bisector of $CE$.