# Circumcenter

There is a neighborhood in Seattle, called the Denny Triangle, because of its triangular shape.

Imagine that you and your two friends live at each vertex of the Denny Triangle. You plan a meeting this weekend at a point that is equidistant from each of your homes.

Do you know that by using perpendicular bisectors to find the circumcenter of this triangle, you can exactly locate your meeting point? Isn't that interesting?

Let's learn more about this!

In this mini-lesson, we will learn all about circumcenter. The journey will take us through properties, interesting facts, and interactive questions on circumcenter.

**Lesson Plan**

**What Is a Circumcenter?**

**Circumcenter Definition**

The circumcenter is the center point of the circumcircle drawn around a polygon. The circumcircle of a polygon is the circle that passes through all of its vertices and the center of that circle is called the circumcenter. All polygons that have circumcircle are known as cyclic polygons.

However, all polygons need not have the circumcircle. Only regular polygons, triangles, rectangles, and right-kites can have the circumcircle and thus the circumcenter.

**The Circumcenter of a Triangle**

All triangles are cyclic and hence, can circumscribe a circle, therefore, every triangle has a circumcenter. The circumcenter of a triangle can be found out as the intersection of the perpendicular bisectors (i.e., the lines that are at right angles to the midpoint of each side) of all sides of the triangle. This means that the perpendicular bisectors of the triangle are concurrent (i.e. meeting at one point).

**Construction of a Circumcenter **

**Circumcenter Geometry**

Steps to construct the circumcenter of a triangle:

**Step 1**: Draw the perpendicular bisectors of all the sides of the triangle using a compass.

**Step 2**: Extend all the perpendicular bisectors to meet at a point. Mark the intersection point as \(\text O \), this is the circumcenter.

**Step 3**: Using a compass and keeping \(\text O \) as the center and any vertex of the triangle as a point on the circumference, draw a circle, this circle is our circumcircle whose center is \(\text O \).

You can construct a circumcenter using the following simulation.

Just place the D end of the compass at the center of the hypotenuse of the triangle and end E at one of the vertex.

Press Draw circle and circumcenter will be drawn by the simulator.

**Properties of a Circumcenter**

Consider any\( \triangle \text {ABC}\) with circumcenter \(\text O \).

1. All the vertices of the triangle are equidistant from the circumcenter.

Join \(\text O \) to the vertices of the triangle.

\[ AO = BO = CO\](radius of the same circle)

Hence, the vertices of the triangle are equidistant from the circumcenter.

2. All the new triangles formed by joining \(\text O \) to the vertices are Isosceles triangles.

3. Angle \( \angle \text {BOC} = 2\angle \text A\) when \( \angle \text A\) is acute or when \(\text O \) and \(\text A\) are on the same side of \(\text {BC}\).

4. Angle \(\angle \text {BOC} = 2( 180^{\circ} - \angle \text A)\) when \( \angle \text A\) is obtuse or \(\text O \) and \(\text A\) are on different sides of \(\text {BC}\).

5. Location for the circumcenter is different for different types of triangles.

\( Triangle \) | \( Location \) | |
---|---|---|

Acute Angle Triangle | Inside the triangle | |

Obtuse Angle Triangle | Outside the triangle | |

Right Angled Triangle | On the Hypotenuse of the triangle |

- For an Equilateral triangle, all the four points (circumcenter, incenter, orthocenter, and centroid) coincide.
- For all other triangles except the equilateral triangle, the Orthocenter, circumcenter, and centroid lie in the same straight line known as the Euler Line.
- Any point on the perpendicular bisector of a line segment is equidistant from the two ends of the line segment.
- Circumcenter divides the equilateral triangle into three equal triangles if joined with vertices of the triangle.

**Locating Circumcenter**

There are various methods through which we can locate the circumcenter \(\text O(x,y)\) of a triangle whose vertices are given as \( \text A(x_1,y_1), \text B(x_2,y_2)\space \text and \space \text C(x_3,y_3)\).

**Method 1: Using the midpoint formula**

**Step 1 :** Calculate the midpoints of the line segments \(\text{AB, AC} \space, and \space \text BC\) using the midpoint formula.

\( \begin{equation} M(x,y) = \left(\dfrac{ x_1 + x_2} { 2} , \dfrac{y_1 + y_2}{2}\right) \end{equation}\) |

**Step 2 : **Calculate the slope of any of the line segments \(\text{AB, AC }\space, and \space \text {BC}\).

**Step 3: **By using the midpoint and the slope of the perpendicular line, find out the equation of the perpendicular bisector line.

\( (y-y_1) = \left(- \dfrac1m \right)(x-x_1)\) |

**Step 4:** Similarly, find out the equation of the other perpendicular bisector line.

**Step 5:** Solve two perpendicular bisector equations to find out the intersection point.

This intersection point will be the circumcenter of the given triangle.

**Method 2: Using the Distance formula**

\(\begin{equation} d = \sqrt{( x - x_1) {^2} + ( y - y_1) {^2}} \end{equation}\) |

**Step 1 :** Find \(d_1, d_2\space and \space d_3\)

\[ \begin{equation} d_1= \sqrt{( x - x_1) {^2} + ( y - y_1) {^2}} \end{equation}\]

\(d_1\) is the distance between circumcenter and vertex \(A\). \[ \begin{equation} d_2= \sqrt{( x - x_2) {^2} + ( y - y_2) {^2}} \end{equation}\] _{ }is the distance between circumcenter and vertex \(B\).

\[ \begin{equation} d_3= \sqrt{( x - x_3) {^2} + ( y - y_3) {^2}} \end{equation}\] \(d_3\) is the distance between circumcenter and vertex \(C\).

**Step 2 : **Now by computing, \(d_1 = d_2\space = \space d_3\)_{ }we can find out the coordinates of the circumcenter.

This is the widely used distance formula to determine the distance between any two points in the coordinate plane.

**Method 3: Using Extended \(\sin\) law**

\(\begin{equation} \dfrac{ a}{ \sin A}=\dfrac{b}{ \sin B} =\dfrac{c} { \sin C} = 2R \end{equation}\) |

Given that \( \text a, \text b \space and \space \text c\) are lengths of the corresponding sides of the triangle and \( \text R\) is the radius of the circumcircle.

By using the extended form of sin law, we can find out the radius of the circumcircle, and using the distance formula can find the exact location of the circumcenter.

**Method 4: Using the Circumcenter formula**

We can quickly find the circumcenter by using the circumcenter of a triangle formula:

\[\begin{equation} O(x, y)=\left(\frac{x_{1} \sin 2 A+x_{2} \sin 2 B+x_{3} \sin 2 C}{\sin 2 A+\sin 2B+\sin 2 C},\\ \frac{y_{1} \sin 2 A+y_{2} \sin 2 B+y_{3} \sin {2} C}{\sin 2 A+\sin 2 B+\sin 2 C}\right) \end{equation}\] |

Where\( \angle \text A, \angle \text B\space and \space \angle \text C\) are respective angles of \( \triangle \text {ABC}\).

- If you move any vertex of the triangle, the perpendicular bisector of the opposite line segment to this vertex will not move. Now, can you say anything about the trajectory of the circumcenter?
- Can the Circumcenter of a triangle be located at any of the vertices of the triangle?

**Solved Examples**

Example 1 |

Shemron has a cake that is shaped like an equilateral triangle of sides \(\sqrt3 \text { inch}\) each. He wants to find out the dimension of the circular base of the cylindrical box which will contain this cake.

**Solution**

Since it is an equilateral triangle, \( \text {AD}\) (perpendicular bisector) will go through the circumcenter \(\text O \). Now using circumcenter facts that the Circumcenter will divide the equilateral triangle into three equal triangles if joined with the vertices.

i.e.

\[\begin{align*} \text {area of } \triangle AOC = \text {area of } \triangle AOB \\= \text {area of } \triangle BOC \end{align*}\]

Therefore,

\[\begin{align*} \text {area of } \triangle {ABC} {} &= 3 \times \text {area of } \triangle BOC \end{align*} \]

Using the formula for the area of an equilateral triangle \[\begin{align*} &= \dfrac{\sqrt3}{4} \times a^2 \end{align*} \]

Also, area of triangle \[\begin{align*} &= \dfrac{1}{2} \times \text { base } \times \text { height } \end{align*} \]

On substituting we get,

\[\begin{align*} {\dfrac{\sqrt3}{4}} \times a^2 &= 3\times \dfrac{1}{2} \times a\times OD\\OD &= \dfrac{1}{2{\sqrt3}} \times a \end{align*}\]

Now for \(\triangle \text{ ABC}\)

Again using formula for area of \(\triangle \text{ ABC}\) = \( \dfrac{1}{2} \times \text { base } \times \text { height } \) = \( \dfrac{\sqrt3}{4} \times a^2 \)

\[\begin{align*}\dfrac {1}2\times a\times (R+OD) &= \dfrac {\sqrt 3}4\times a^2 \\\dfrac12 a\times \left( R+\dfrac a{2\sqrt3}\right) &= \dfrac{\sqrt3}4\times a^2\\R &= \dfrac a{\sqrt3} \end{align*}\]

Substituting,

\[ \begin{align*}a & = \sqrt3 \end{align*}\]

\[\text {R} = 1 \space inch\] |

Example 2 |

Charlie came to know that the circumcenter of a Right-angled triangle lies in the exact center of its hypotenuse. He wants to check this with a Right-angled triangle of sides \(\text L(0,5), \text M(0,0)\space and\space \text N(5,0)\). Can you help him in confirming this fact?

**Solution**

Using the circumcenter formula or circumcenter of a triangle formula from circumcenter geometry:

\[ \begin{equation} O(x, y)=\left(\dfrac{x_{1} \sin 2 A+x_{2} \sin 2 B+x_{3} \sin 2 C}{\sin 2 A+\sin 2B+\sin 2 C},\\ \dfrac{y_{1} \sin 2 A+y_{2} \sin 2 B+y_{3} \sin {2} C}{\sin 2 A+\sin 2 B+\sin 2 C}\right) \end{equation}\]

Putting the corresponding values,

\[O(x,y) = \dfrac { (0 + 0 + 5 \times 1)}{ (0 + 1 + 1) }, \dfrac { (5 \times 1 + 0 + 0)}{(0 + 1 + 1)}\]

\[ O(x,y) = \dfrac {5}{2} , \dfrac {5}{2}\] |

Example 3 |

Thomas has triangular cardboard whose one side is \(19 \text { inch}\) and the opposite angle to that side is \(30^{\circ}\). He wants to know the base area of the cylindrical box so that he can fit this card in it completely.

**Solution**

Using Extended \( \sin\) law,

\[\begin{equation} \dfrac{ a}{ \sin A}=\dfrac{b}{ \sin B} =\dfrac{c} { \sin C} = 2R \end{equation}\]

\[\dfrac{19} { \sin30} = 2R\]

\[R = 19 \text { in}\]

Now, the area of the circumcircle:

\[\pi r^2 = \pi \times{19^2}\]

\[\text{ Area} = 1133.54 \space \text { in}^2\] |

Example 4 |

The perpendicular bisectors of the triangle intersect at \( \text O \). Find out the length of \( \text YO \).

**Solution**

We know that for any triangle, its circumcenter is equidistant from its vertices.

Which means,

\[ \text {XO } = \text { YO } = \text { ZO } \]

Hence,

\[ \text {YO } = 18\] |

Example 5 |

In the figure, the Right-angled triangle is shown with \( \text A \) at the origin. The length of one side of the triangle is \( 5 \text { in} \) and the coordinate of the circumcenter is \( \text O (2.5,6) \). Find the length of the hypotenuse of the triangle.

**Solution**

Given,

Coordinates of circumcenter \[ \text O (2.5,6) \].

Length of side, \[ \text { AB } = 5 \text { in} \].

Using the circumcenter property, that, for a right-angled triangle, the circumcenter lies at the midpoint of the hypotenuse.

Also, the circumcenter lies at the bisector of all sides which means,

\[ \text { AD } = \text {DC } \]

So, coordinates of \( \text D \) will be \( ( 0, 6) \).

Therefore, coordinates of C will be \( ( 0, 12) \).

Now, as the length of \( \text { AC } \) is \( 12 \) and \( \text { AB } \) is \( 5 \), by using Pythagoras theorem we can find BC.

\[ 5^2 + 12^2 = x^2 \]

\[ BC = \sqrt {169} \]

\[ BC = 13 \]

\[ \therefore\ \text {Hypotenuse } = 13 \text{ inch}\] |

**Interactive Questions**

**Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.**

**Let's Summarize**

We hope you enjoyed learning about the circumcenter with the simulations and interactive questions. Now, you will be able to easily solve problems on the circumcenter and its properties in math.

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**Frequently asked questions - FAQS**

## 1. How to Find the Circumcenter of a Triangle?

We can find circumcenter by using the circumcenter of a triangle formula, where the location of the circumcenter is \(\text O(x,y)\) and the coordinates of a triangle are given as \( \text A(x_1,y_1), \text B(x_2,y_2)\space \text and \space \text C(x_3,y_3)\).

\[\begin{equation} O(x, y)=\left(\dfrac{x_{1} \sin 2 A+x_{2} \sin 2 B+x_{3} \sin 2 C}{\sin 2 A+\sin 2B+\sin 2 C}\right),\\ \left(\dfrac{y_{1} \sin 2 A+y_{2} \sin 2 B+y_{3} \sin {2} C}{\sin 2 A+\sin 2 B+\sin 2 C}\right) \end{equation}\]

Where \(A\), \(B\) ,and \(C\) are the respective angles of the triangle.

## 2. How do you find the Circumcenter with vertices?

Using the Distance formula, where the vertices of the triangle are given as \( A(x_1,y_1),B(x_2,y_2)\space \text and \space C(x_3,y_3)\) and the coordinate of the circumcenter is \(O(x,y)\).

Find d_{1}, d_{2,} and d_{3} by using following formlae

\[\begin{equation} d_1 = \sqrt{( x - x_1) {^2} + ( y - y_1) {^2}} \end{equation}\] \( d_1 \) is the distance between circumcenter and vertex \(A\).

\[\begin{equation} d_2 = \sqrt{( x - x_2) {^2} + ( y - y_2) {^2}} \end{equation}\] \( d_2 \)_{ }is the distance between circumcenter and vertex \(B\).

\[\begin{equation} d_3 = \sqrt{( x - x_3) {^2} + ( y - y_3) {^2}} \end{equation}\] \( d_3 \) is the distance between circumcenter and vertex \(C\).

now, by computing \(d_1 = d_2 = d_3\)_{ , }we can find out the coordinates of the circumcenter.

## 3. What is the Difference Between a Circumcenter and an Incenter?

The incenter is the center of the circle inscribed inside a triangle (incircle) and the circumcenter is the center of a circle drawn outside a triangle (circumcircle). The incenter can never lie outside the triangle, whereas, the circumcenter can lie outside of the triangle.

## 4. Does Every Triangle Have a Circumcenter?

Yes, as all the triangles are cyclic in nature which means that they can circumscribe a circle, and hence, every triangle has a circumcenter.

## 5. Are Circumcenter and Centroid the Same?

Except for Equilateral triangles, the circumcenter and centroid are two distinct points as they do not coincide with each other.