Finding Center of Circle

Theorem

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

Geometric Proof

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 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 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.

$\blacksquare$

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.

Historical Note

This is Proposition 1 of Book III of Euclid's The Elements.

Alternative Geometric 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 Geometric 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.