# Circumcenter of a triangle

The **circumcenter** of a triangle (*O*) is the point where the three perpendicular bisectors (M_{a}, M_{b} y M_{c}) of the sides of the triangle intersect. It can be also defined as one of a triangle’s points of concurrency.

The perpendicular bisector of a triangle is a line perpendicular to the side that passes through its midpoint.

The **circumcenter** (*O*) is the central point that forms the origin of the **circumcircle** (circumscribed circle) in which all three vertices of the triangle lie on the circle.

It’s possible to find the radius (*R*) of the circumcircle if we know the three sides and the semiperimeter of the triangle.

The radius of the circumcircle is also called the **triangle’s circumradius**.

The formula for the circumradius is:

Download this **calculator** to get the results of the formulas on this page. Choose the initial data and enter it in the upper left box. For results, press ENTER.

Triangle-total.rar or Triangle-total.exe

Note. Courtesy of the author: **José María Pareja Marcano**. Chemist. Seville, Spain.

**Where is the Circumcenter of a Triangle Located?**

- If it’s an obtuse triangle the circumcenter is located
**outside**the triangle (as we see in the picture above). - If it’s an acute triangle the circumcenter is located
**inside**the triangle. - If it’s a right triangle the circumcenter lies on the
**midpoint**of the hypotenuse (the longest side of the triangle, that is opposite to the right angle (90°). We can see an example in the figure below.

See the Thales’ Theorem.

## Euler Line

In any **non-equilateral triangle** the orthocenter (*H*), the centroid (*G*) and the **circumcenter** (*O*) are aligned. The line that contains these three points is called the Euler Line.

In an equilateral triangle all three centers are in the same place.

The relative distances between the triangle centers remain constant.

**Distances between centers**:

It is true that the distance from the orthocenter (*H*) to the centroid (*G*) is twice that of the centroid (*G*) to the **circumcenter** (*O*). Or put another way, the *HG* segment is twice the *GO* segment:

When the triangle is equilateral, the centroid, orthocenter, circumcenter, and incenter coincide in the same interior point, which is at the same distance from the three vertices.

This distance to the three vertices of an equilateral triangle is equal to from one side and, therefore, to the vertex, being *h* its altitude (or height).

## Exercise 1

Find the radius *R* of the circumscribed circle (or **circumcircle**) of a triangle of sides *a* = 9 cm, *b* = 7 cm and *c* = 6 cm.

**Solution:**

We get the semiperimeter *s*:

We apply the formula for the radius *R* of the circumscribed circle, giving the following values:

So, the radius R is 4.5 cm.

## Exercise 2

Find the coordinates of the **circumcenter of a triangle** *O* *ABC* whose vertices are *A*(3, 5), *B*(4, -1) y *C*(-4, 1).

**SOLUTION:**

In order to find the circumcenter O we have to solve the equations for two perpendicular bisectors *M _{a}* (perpendicular to side

*a*) and

*M*(perpendicular to side

_{b}*b*) and see where is located the intersection point (that is the circumcenter

*O*) of both perpendicular bisectors.

**STEP 1: Find the equation for the perpendicular bisector M_{a}.**

Firstly we will find the equation of the line that passes through side *a*, which is the opposite of vertex *A*. This equation is obtained knowing that it passes through points *B* (4, -1) and *C* (-4, 1). The general equation of the line that passes through two known points is:

The equation of the line that contains side *BC* and its slope *m* will be:

So, the slope *m* for the line a is -1/4.

Now, we get the coordinates of the midpoint *r* between vertices *B* and *C*, i.e. midpoint of side *a*.

Where is r?

So,

The slope of the line that contains the perpendicular bisector *M _{a}*, being perpendicular to the side

*a*, is the inverse and of the opposite sign to the slope of the line found that contains side

*a*. So, we have that:

So, the slope of the line *M _{a}* is 4 because the slope of the line a it was -1/4.

Since we know that perpendicular bisector *M _{a}* passes through the midpoint

*r*(located at (0, 0)) and we know its slope

*m*, which is equal to 4, now we can obtain the equation for the line

_{p}*M*:

_{a} This is the equation for the perpendicular bisector *M _{a}*.

**STEP 2: Find the equation for the perpendicular bisector M_{b}.**

Now we proceed in the same way to find the equation of the line that contains the perpendicular bisector *M _{b}*, that is, the one that passes through the midpoint

*s*and is perpendicular to the side

*b*between vertices

*A*and

*C*.

First, we calculate the slope of the line *b* (or side *b*):

Then we find the midpoint *s* coordinates between vertices *A* and *C*:

The equation of the line that contains the perpendicular bisector *M _{b}*, that is, the one starting from the midpoint

*s*is perpendicular to side

*b*. Therefore, the slope of this line will therefore be –7/4 (inverse and of the opposite sign). With the slope of a line and one of its points we can find the equation:

We have the equations of two of the perpendicular bisectors of the triangle, *M _{a}* and

*M*:

_{b}Next, we solve this system of two equations in two variables using the substitution method, the most suitable, given the form of the first equation:

Finally, we have that *x* = 0,37 and *y* = 1,48.

Hence, circumcenter *O* (0.37, 1.48).

The circumcenter *O* is the centerpoint of the circumscribed circle: