Finding Center of Circle

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

Geometric Proof

 * Euclid-III-1.png

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

Bisect $BC$ at $$G$$.

Construct $DE$ perpendicular to $$BC$$ at $$G$$, where $$D$$ and $$E$$ are where this perpendicular meets the circle.

Bisect $DE$ at $$A$$.

Then $$A$$ is the center of the circle.

The proof is as follows.

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

Join $$FB, FG, FC$$.

As $$F$$ is the center, then $$FB = FC$$.

Also, we have $$GB = GC$$ as $$G$$ bisects $$BC$$.

So from Triangle Side-Side-Side Equality, $$\triangle FGB = \triangle FGC$$.

Hence $$\angle BGF = \angle CGF$$.

But from book 1 definition 10:
 * "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 $$\angle FGC$$ is a right angle.

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

So $$\angle FGC = \angle AGC$$, and this can happen only if $$F$$ lies on $$DE$$.

But if $$F$$ is on DE, then as $$F$$ is, as we suppose, at the center of the circle, then $$FD = FE$$, and so $$F$$ bisects $$DE$$.

But then $$FD = AD$$, and so $$F = A$$.

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."

Augustus De Morgan preferred to prove this first, wording it as:
 * "The line which bisects a chord perpendicularly must contain the center"

and then use that to prove this.

See Perpendicular Bisector of Chord Passes Through Center.