Finding Center of Circle

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

Proof

 * Euclid-III-1.png

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 :

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 if $G$ is on $CE$, then as $G$ is, as we suppose, at the center of the circle, then $GC = GE$, and so $G$ bisects $CE$.

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

Hence the result.

Porism
From this result, Euclid derived the following porism:


 * If in a circle a straight line cut a straight line into two equal parts and at right angles, the center of the circle is on the cutting straight line.

Alternative Proof
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, the bisector of $CE$.

Note on Alternative Proof
This proof was formulated by Augustus De Morgan who preferred to prove the more fundamental result first, wording it as:
 * The line which bisects a chord perpendicularly must contain the center

and then use that to prove this.