The circumcenter of a triangle can be found using five different methods. Three methods consist in using different algebraic and geometric techniques to determine the equations of the perpendicular bisectors and determine the point of intersection. The fourth method consists in using a formula with the coordinates of the vertices and the measures of the angles. The fifth method consists in determining the circumcenter graphically.

In this article, we will learn how to find the circumcenter of a triangle using five different methods. We will solve some practice problems.

## What is the circumcenter of a triangle?

The circumcenter is the point of intersection of the perpendicular bisectors of the sides of the triangle. In the diagram below, we can see that point O is the circumcenter:

Remember that the perpendicular bisectors are the perpendicular segments that pass through the midpoints of each side of the triangle.

Alternatively, we can define the circumcenter as the center of the circumscribed circle, which passes through all three vertices of the triangle.

## Circumcenter of common triangles

The location of the circumcenter varies depending on the type of triangle we have.

### Circumcenter of an acute triangle

The circumcenter of all acute triangles always lies inside the triangle.

### Circumcenter of an obtuse triangle

The circumcenter of all obtuse triangles always lies outside the triangle.

### Circumcenter of a right triangle

The circumcenter of all right triangles is located on the hypotenuse of the right triangle. Also, the hypotenuse of the right triangle corresponds to the diameter of the circumscribed circle.

### Circumcenter of an equilateral triangle

The circumcenter, orthocenter, incenter, and centroid of all equilateral triangles are located in the same position.

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## Finding the circumcenter of a triangle graphically

To find the circumcenter of a triangle graphically, we have to sketch the perpendicular bisectors and find the point of intersection.

We can use a compass to determine two perpendicular bisectors by following these steps:

**Step 1:** Draw an arc on side AB using vertex B as the center and a radius with a length slightly greater than half of AB. For example, if AB is 4 units, we can use a radius of 2.5.

** Step 2:** Use vertex C as the center with the same radius as in step 1 and draw an arc on side AB.

** Step 3:** Draw a line that passes through points D and E. This is a perpendicular bisector.

** Step 4:** Follow a similar process to draw intersecting arcs from C and A to form points F and G and draw a line through those points.

** Step 5:** Mark the point of intersection of the drawn lines.

The lines drawn are the perpendicular bisectors since they are perpendicular to their corresponding sides and pass through the midpoint of the sides. Therefore, the point of intersection is the circumcenter of the triangle.

## Finding the circumcenter of a triangle algebraically

We can find the circumcenter of a triangle using four main methods. We are going to use the following triangle that has the vertices $latex A=(x_{1},~y_{1})$, $latex B=(x_{2},~y_{2})$ and $latex C=(x_{3},~y_ {3})$. The circumcenter is located at (*x, y*).

### Method 1: Using the distance formula

** Step 1:** Using the distance formula, $latex d=\sqrt{(x-x_{1})^2+(y-y_{1})^2}$, we can find $latex d_{1}, ~d_{ 2}$ and $latex d_{3}$, which are the distances from the circumcenter to vertices A, B and C respectively:

$latex d_{1}=\sqrt{(x-x_{1})^2+(y-y_{1})^2}$

$latex d_{2}=\sqrt{(x-x_{2})^2+(y-y_{2})^2}$

$latex d_{3}=\sqrt{(x-x_{3})^2+(y-y_{3})^2}$

** Step 2:** We will obtain three equations in terms of the coordinates of the circumcenter (

*x, y*). We use two of these equations and any method of solving systems of equations to obtain a solution (

*x, y*). The solution to the system is the coordinates of the circumcenter.

### Method 2: Using the midpoint formula

** Step 1:** Using the midpoint formula, $latex M(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$, we calculate the midpoints of the three sides of the triangle AB, BC, AC.

** Step 2:** Find the slopes of the triangle sides AB, BC and AC using the slope formula:

$latex m_{1}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$

$latex m_{2}=\frac{y_{3}-y_{2}}{x_{3}-x_{2}}$

$latex m_{3}=\frac{y_{3}-y_{1}}{x_{3}-x_{1}}$

where, $latex m_{1}$ is the slope of AB, $latex m_{2}$ is the slope of BC, and $latex m_{3}$ is the slope of AC.

** Step 3:** Use the coordinates of the midpoint and the slope of each side of the triangle to obtain an equation for its perpendicular bisector (perpendicular line).

$latex (y-y_{1})=-\frac{1}{m}(x-x_{1})$

** Step 4:** Use the equations of two perpendicular bisectors of the triangle to form a system of equations. The solution to the system of equations is the coordinates of the circumcenter.

### Method 3: Using the extended law of sines

** Step 1:** Using the extended law of sines, we can obtain the length of the radius of the circumscribed circle. The extended law of sines is written as follows:

$latex \frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}=2R$

where, *a, b, c* are the lengths of the sides of the triangle and *R* is the radius of the circumscribed circle.

** Step 2:** Use the distance formula, along with the coordinates of the vertices to obtain two equations in terms of the coordinates of the circumcenter.

** Step 3:** Solve with any method of systems of equations and we will obtain the coordinates of the circumcenter.

### Method 4: Using the circumcenter formula

If we know the coordinates of the three vertices and the size of the three angles, we can use the circumcenter formula to find the coordinates of the circumcenter quickly:

$$O(x, y)=\left(\frac{x_{1}\sin(2A)+x_{2}\sin(2B)+x_{3}\sin(2C)}{\sin(2A)+\sin(2B)+\sin(2C)}, ~\frac{y_{1}\sin(2A)+y_{2}\sin(2B)+y_{3}\sin(2C)}{\sin(2A)+\sin(2B)+\sin(2C)}\right)$$

where, $latex A(x_{1}, y_{1})$, $latex B(x_{2}, y_{2})$, $latex C(x_{3}, y_{3})$ are the vertices of the triangle and A, B, C are the corresponding angles.

## Circumcenter of a triangle – Examples with answers

In the following examples, we apply what we have learned about the circumcenter of a triangle.

### EXAMPLE 1

Use the midpoint formula to find the coordinates of the circumcenter of the following triangle.

##### Solution

We use the midpoint formula to find the coordinates of the midpoints of the sides of the triangle:

- $latex (x_{1},~y_{1})=(2, ~4)$
- $latex (x_{2},~y_{2})=(1,~1)$
- $latex (x_{3},~y_{3})=(5,~1)$

Midpoint of AB:

$latex M_{1}=(\frac{2+1}{2}, \frac{4+1}{2})$

$latex M_{1}=(\frac{3}{2}, \frac{5}{2})$

Midpoint of AC:

$latex M_{3}=(\frac{2+5}{2}, \frac{4+1}{2})$

$latex M_{3}=(\frac{7}{2}, \frac{5}{2})$

Now, we have to determine the slopes of AB and AC.

Slope of AB:

$latex m_{AB}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$

$latex m_{AB}=\frac{4-1}{2-1}$

$latex m_{AB}=3$

Slope of AC:

$latex m_{BC}=\frac{y_{3}-y_{1}}{x_{3}-x_{1}}$

$latex m_{BC}=\frac{1-4}{5-2}$

$latex m_{BC}=-1$

Now, we have to determine the equations of the lines perpendicular to the sides and that pass through the midpoints found.

$latex (y-y_{1})=-\frac{1}{m}(x-x_{1})$ (1)

$latex (y-\frac{5}{2})=-\frac{1}{3}(x-\frac{3}{2})$ (1)

$latex 6y-15=-2x+3$

$latex 6y+2x=18$

$latex 3y+x=9$

$latex (y-y_{1})=-\frac{1}{m}(x-x_{1})$ (2)

$latex (y-\frac{5}{2})=-\frac{1}{-1}(x-\frac{7}{2})$ (2)

$latex 2y-5=2x-7$

$latex 2y-2x=-2$

$latex y-x=-1$

Using any method to solve the system of equations, we get *x*=3, *y*=2, which are the coordinates of the circumcenter.

### EXAMPLE 2

The circle passes through the three vertices of the following triangle, so it is circumscribed. What is its area?

##### Solution

We can find the radius of the circumscribed circle using the extended law of sines:

$latex \frac{a}{\sin (A)}=\frac{b}{\sin (B)}=\frac{c}{\sin (C)}=2R$

$latex \frac{20}{\sin (30)}=2R$

$latex R=20$ cm

Now, we can find the area of the circle easily:

$latex A=\pi r^2$

$latex A=\pi (20)^2$

$latex A=400\pi$ cm²

### EXAMPLE 3

What are the coordinates of the circumcenter of the following isosceles right triangle?

##### Solution

We can use the circumcenter formula since we have the coordinates of the vertices and the measure of the angles.

The triangle is an isosceles right triangle, which means it has one 90° angle at A and two 45° angles at B and C. Also, we know that sine of 90° equals 1 and sine of 180° equals 0. Therefore, we have:

$$O(x, y)=\left(\frac{x_{1}\sin(2A)+x_{2}\sin(2B)+x_{3}\sin(2C)}{\sin(2A)+\sin(2B)+\sin(2C)}, ~ \frac{y_{1}\sin(2A)+y_{2}\sin(2B)+y_{3}\sin(2C)}{\sin(2A)+\sin(2B)+\sin(2C)}\right)$$

$latex O(x, y)=(\frac{(0+0+4 \times 1}{0+1+1},~\frac{0+4\times 1+0}{0+1+1})$

$latex O(x, y)=(\frac{4}{2},~\frac{4}{2})$

$latex O(x, y)=(2,~2)$

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## Circumcenter of a triangle – Practice problems

Solve the following practice problems using an appropriate method to obtain information about the circumcenter of a triangle.

## See also

Interested in learning more about the incenter, orthocenter, centroid, and circumcenter of a triangle? Take a look at these pages:

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