# Circumcenter of a Triangle – Definition, Formula and Examples

The circumcenter of a triangle is the point where the three perpendicular bisectors of the triangle intersect. The circumcenter can also be defined as the center of the circumscribed circle that passes through the three vertices of the triangle. The location of the circumcenter varies depending on the type of triangle we have.

Here, we will learn more details about the circumcenter of a triangle using diagrams. We will learn about its formula and how to find it graphically.

##### GEOMETRY

Relevant for

Learning about the circumcenter of a triangle.

See definition

##### GEOMETRY

Relevant for

Learning about the circumcenter of a triangle.

See definition

## Definition of the circumcenter of a triangle

The circumcenter is the point where the perpendicular bisectors of the three sides of the triangle intersect.

Remember that the perpendicular bisectors are the perpendicular segments that pass through the midpoint of another segment. In this case, the perpendicular bisectors of a triangle pass through the midpoints of all three sides.

Alternatively, the circumcenter of a triangle can be defined as the center of the circumscribed circle. In turn, a circumscribed circle is a circle that passes through all the vertices of a polygon.

The location of the circumcenter varies depending on the type of triangle. For example, for equilateral triangles, the circumcenter is located in the same position as the centroid. However, for obtuse triangles, the circumcenter lies outside the triangle.

### Circumcenter of an acute triangle

All acute triangles have the circumcenter located inside the triangle. Recall that in an acute triangle, all internal angles measure less than 90°.

### Circumcenter of an obtuse triangle

All obtuse triangles have the circumcenter located outside the triangle. Recall that in an obtuse triangle, one internal angle is greater than 90°.

### Circumcenter of a right triangle

All right triangles have the circumcenter located on the hypotenuse of the triangle. Remember that a right triangle is characterized by having an angle of 90°.

### Circumcenter of an equilateral triangle

For all equilateral triangles, the circumcenter, orthocenter, centroid, and incenter are located in the same position. Remember that these triangles have all their sides of the same length.

## Formula for the circumcenter of a triangle

The circumcenter of a triangle can be found using four main formulas. With these formulas, we will be able to locate the coordinates (x, y) of the circumcenter of a triangle that has the vertices $latex A=(x_{1},~y_{1})$, $latex B=(x_{2},~y_{2})$ and $latex C=(x_{3},~y_{3})$.

### Formula 1: Distance formula

Remember that the distance formula is $latex d=\sqrt{(x-x_{1})^2+(y-y_{1})^2}$. Therefore, we can find $latex d_{1}, ~d_{2}$ and $latex d_{3}$:

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

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

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

where, $latex d_{1}$ is the distance from the circumcenter to vertex A, $latex d_{2}$ is the distance from the circumcenter to vertex B, and $latex d_{3}$ is the distance from the circumcenter to vertex C.

With this, we will obtain three equations in terms of the coordinates of the circumcenter (x, y). We can take two of these equations and solve them using any method of solving systems of equations. The solution to the system is the coordinates of the circumcenter.

### Formula 2: Midpoint formula

Recall that the midpoint formula is $latex M(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$. Therefore, we can use this formula to find the midpoints of the three sides of the triangle AB, BC, AC.

Then, we use the slope formula to find the slopes of AB, BC, and AC:

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

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

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

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.

Using the coordinates of the midpoint and the slope of each side of the triangle, we can obtain an equation for its perpendicular bisector.

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

Using the equations of the perpendicular bisectors of two sides of the triangle, we can form a system of equations and solve with any method. The solution to the system of equations will be the point of intersection of the perpendicular bisectors, that is, the coordinates of the circumcenter.

### Formula 3: Extended law of sines

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.

Using this formula, we can get the length of the radius of the circle. Then, we can use the distance formula and using the coordinates of the vertices, we will obtain two equations.

The equations are solved using any method of systems of equations and we will obtain the coordinates of the circumcenter.

### Formula 4: Circumcenter formula

If we know the coordinates of the three vertices and the sizes 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.

## Finding the circumcenter of a triangle graphically

The circumcenter of a triangle can be found graphically by plotting the perpendicular bisectors and finding the point of intersection. This is extremely useful when we want to find the circumcenter, but we don’t have the coordinates of the vertices of the triangle.

We have to determine two perpendicular bisectors to find the point of intersection. So, we follow the following steps using a compass:

Step 1: Use vertex B as the center and use 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. With that radius, draw an arc on side AB.

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

Step 3: We got two intersection points, D and E. Draw a line that passes through those points.

Step 4: Draw intersecting arcs from C and A using a radius that is slightly larger than half CA to form points F and G and draw a line through those points.

Step 5: Find the point of intersection of the drawn lines.

The lines drawn are perpendicular to their corresponding sides. Also, the lines pass through the midpoint of the sides. This means that they are two perpendicular bisectors of the triangle. Therefore, the point of intersection is the circumcenter of the triangle.

## Circumcenter of a triangle – Examples with answers

In the following examples, we use different methods to find the coordinates of the circumcenter of a triangle.

### EXAMPLE 1

Determine the circumcenter of the following triangle using the midpoint formula.

Solution: We need to use the midpoint formula to find the coordinates of the midpoints of the sides of the triangle and then we find the slope of line that goes through those points. We use the following vertices:

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

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 find 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 find the equations of the perpendicular lines passing 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)

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

$latex 2y-2x=-2$

$latex y-x=-1$

Solving the system of equations, we get x=3, y=2, which are the coordinates of the circumcenter.

### EXAMPLE 2

Find the area of the circumscribed circle that passes through the three vertices of the following triangle.

Solution: We can use the extended law of sines to find the radius of the circumscribed circle:

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

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

$latex R=20$ cm

Using that radius, we can easily find the area of the circle:

$latex A=\pi r^2$

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

$latex A=400\pi$ cm²

### EXAMPLE 3

Use the circumcenter formula to determine the coordinates of the circumcenter of the following isosceles right triangle.

Solution: Since the triangle is an isosceles right triangle, we know that 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° is equal to 0. Therefore, we have:

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

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

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

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