Circumcenter and Circumcircle

The circumcircle is a circle that, for a triangle, has all three vertices on its circumference. The circumcenter of a triangle is the center of this circle.

But how do we know a circumcenter exists for every nondegenerate triangle?

Draw the perpendicular bisectors of $\overline{AB}$ and $\overline{BC}$. They will intersect at a point; call it $O$. (Intuitively, you can see that if they don't intersect, the triangle will be degenerate. We won't prove this here, but we will leave it to you as an exercise.)

Since $O$ lies on perpendicular bisectors, we have $\overline{OA} \cong \overline{OB}$ and $\overline{OB} \cong \overline{OC}$, or $\overline{OA} \cong \overline{OB} \cong \overline{OC}$. Therefore, $O$ is equidistant from each of the three vertices of the triangle. But the locus of points a given distance away from a point is a circle! So we can draw a circle through the points, where $O$ is the center:

We're going to set out today to show that for every triangle $ABC$, there is a circle that goes through all 3 points! This is called the circumcircle. To find the circle, it's best to start by finding the center, the circumcenter, which is normally called $O$. Since $O$ is the center of the circle, $OA = OB = OC$. Let's find $O$!

Here's the triangle $ABC$ to look at.

Good! The points $X$ that satisfy $XB = XC$ are exactly those that are on the perpendicular bisector of $BC$, or the line perpendicular to $BC$ that passes through the midpoint.

A circle centered at $B$ through the midpoint of $AB$.

@ Draw the circle out! Does it look like every point $Y$ on this circle satisfies $YA = YB$?

Once again, these points are the exactly those on the perpendicular bisector of $AB$.

$O$ is on both perpendicular bisectors because $OA = OB$ AND $OB = OC$. So $O$ is on both lines! So $O$ is exactly where the lines intersect!

We've used the sides $AB$ and $BC$ to draw perpendicular bisectors, but not $AC$ yet! But I claim that in fact, $O$ also lies on the perpendicular bisector of $AC$. That is, all 3 perpendicular bisectors pass through a single point!

Why does $O$ pass through the perpendicular bisector of $AC$ also? When you think you have an answer, continue.

This is because $OA = OB$, and $OB = OC$, so $OA = OC$ also! Since $OA = OC$, it also lies on the perpendicular bisector of $AC$. Finally, since $OA = OB = OC$, there is a circle centered at $O$ passing through $A, B, C$. This is how you find the circumcenter and circumcircle in a triangle.