Finding Center of Circle

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

In the words of Euclid:

To find the centre of a given circle.

(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:

$\triangle ADG = \triangle BDG$


$\angle ADG = \angle BDG$

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.



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 $CE$.