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# Circumcenter|Definition & Meaning

## Definition

The **circumcenter** is the middle point of the **circumcircle** tempted around a polygon. The **polygon’s** circumcircle is the circle that **crosses** through all of its vertices and the **middle** of that circle is named the **circumcenter.** All polygons which have **circumcircles** are comprehended as **cyclic polygons.** Only standard polygons like **triangles** and rectangles can have a **circumcircle.**

The **circumcircle** of a polygon in geometry is a circle that passes through all the **polygon’s** vertices. The center of this **circle** is known as the **circumcenter** and its radius is named as the **circumradius**.

The **circumcenter** of a triangle is **described** as the juncture where the **perpendicular** bisectors of the sides of the **triangle** cross. The **circumcenter** is the point of **concurrence** of the bisector of the **triangle’s** sides. It is **characterized** by **P(X, Y)**. The circumcenter is **likewise** the middle of the **circumcircle** of that triangle. The **circumcenter** of the triangle can either be **inside** or **outside** the triangle.

## What is the Circumcenter of Triangle

The **circumcenter** in geometry is the **juncture** at which the three **perpendicular** bisectors of the **triangle’s** side **meet,** and the **length** from each of the **three** triangle **vertices** is exactly the **same.**

The **circumcenter** of a triangle is the point where three **perpendicular** bisectors from the **triangle’s** sides **bisect** or **meet.** The **circumcenter** of a triangle is also called the point of **concurrence** of a triangle. The juncture of origin of a **circumcircle** is a circle **etched** inside a triangle also **named** as the **circumcenter.**

The **circumcenter** of the triangle can be **figured** out as the **conjunction** of the perpendicular **bisectors** that is the lines that are at right angles to the **middle** of each side of all flanks of the triangle. This **indicates** that the perpendicular bisectors of the **triangle** are concurrent which means that they **meet** at one point. All triangles are cyclic and thus, can **circumscribe** a circle, consequently, a circumcenter is found in every **triangle.** To produce the circumcenter of a triangle, **perpendicular** bisectors of any two flanks of a triangle are **marked.**

## Circumcenter and Circumcircle

The circle **cannot** be circumscribed in every **polygon.** A cyclic polygon is a **polygon** that does have one. cyclic polygon is sometimes called as a **concyclic** polygon because of its **concyclic** vertices. All triangles, all rectangles, all **regular** simple polygons, all **right kites,** and all **isosceles trapezoids** are **cyclic.**

An associated idea is the one of a **lowest** bounding circle, which is the most **undersized** circle that fully **contains** the polygon inside of it if the **center** of the circle is within the **polygon.** Every **polygon** has a **distinctive** lowest bounding circle, which may be **formed** by a linear time **algorithm.** Actually, if a circle is **circumscribed** within the **polygon** then it might be **distinct** from

its minimum **bounding** circle. For instance, for an **obtuse** triangle, the lowest bounding **circle** has the most **extended** side as a **diameter** and does not cross through the **opposing** vertex.

## Properties of Circumcenter

Following are **some** of the properties of the **circumcenter** of the triangle:

- The
**circumcenter**lies in the center of the**circumcircle.** - All the
**triangle’s**vertices are**equidistant**from the circumcenter. - The
**circumcenter**lies inside the triangle in an acute**triangle.** - The
**circumcenter**lies outside of the triangle in an**obtuse**triangle.

## Method to Calculate the Circumcenter of a Triangle

**Following** are steps to find the **circumcenter** of a triangle:

**Compute**the midpoint of given coordinates, for example,**midpoints**of**AB, BC,**and**AC.**- Compute the
**slope**of the respective line. - By utilizing the
**midpoint**and the slope, figure out the**equation**of the line which is y – y_{1}=m(x – x_{1}) - figure out the
**equation**of the other line in the same**way.** **Compute**two bisector equations by**estimating**t the intersection**point**- The
**calculated**meeting point will be the**circumcenter**of the provided triangle.

## Finding Circumcenter Using Linear Equations

The **circumcenter** can be estimated by **constructing** linear equations utilizing the distance formula. Let us consider **circumcenter** has the coordinates of (X, Y). According to the **properties** of the circumcenter , the length of (X, Y) from each vertex of a triangle will be the **exact** same.

Assume that D_{1} is the **length** between circumcenter (X, Y) and the vertex is (x_{1}, y_{1}), then the formula is **presented** by,

D_{1} = $\sqrt{(X-x_1)^2 + (Y-y_1)^2}$

D_{2} = $\sqrt{(X-x_2)^2 + (Y-y_2)^2}$

D_{3} = $\sqrt{(X-x_3)^2 + (Y-y_3)^2}$

Now, since D_{1}=D_{2} and D_{2}=D_{3}, we get

\[ \sqrt{ (X-x_1)^2 + (Y-y_1)^2} = \sqrt{ (X-x_2)^2 + (Y-y_2)^2} \]

Two linear **equations** are acquired from this and by **simplifying** the linear equations by **utilizing** the elimination or **substitution** method, the **coordinates** of the circumcenter can be **acquired.**

## An Example of Finding the Circumcenter

The **vertices** of a triangle **ABC** are given as A = (2, 2), B = (0, 4), and C = (4, 4), find the coordinates of its **circumcenter** ?

### Solution

Suppose P(x, y) to be the **coordinates** of the **circumcenter,**

D_{1} be the **length** from the **vertex A** to the circumcenter,

D_{2} be the **length** from the **vertex B** to the circumcenter,

and D_{3} be the **length** from the **vertex C** to the circumcenter.

Given the **vertexes** are: A(x_{1}, x_{1}) = A(2, 2), B(x_{2}, x_{2}) = B(0, 4), and C(x_{3}, x_{3}) = C(4, 4).

Using the euclidean **distance** formula:

D_{1} = $\sqrt{(X-x_1)^2 + (Y-y_1)^2}$

D_{2} = $\sqrt{(X-x_2)^2 + (Y-y_2)^2}$

D_{3} = $\sqrt{(X-x_3)^2 + (Y-y_3)^2}$

Since D_{1}= D_{2} = D_{3}.

D_{1} = D_{2} gives:

\[ \sqrt{(X-2)^2 + (Y-2)^2} = \sqrt{(X-0)^2 + (Y-4)^2} \]

Squaring the above gives the following:

x^{2} + 4 – 4x + y^{2} + 4 – 4y = x^{2} + y^{2} + 16 – 8y

-4x – 4y + 8 = 16 – 8y

-4x – 8 + 4y = 0

4y – 4x = 8

-x + y = 2

D_{1} = D_{3 }gives:

\[ \sqrt{(X-2)^2 + (Y-2)^2} = \sqrt{(X-4)^2 + (Y-4)^2} \]

Squaring both sides, we get:

x^{2} + 4 – 4x + y^{2} + 4 – 4y = x^{2} + 16 – 8x + y^{2} + 16 – 8y

8 – 4x – 4y = 32 – 8x – 8y

4x + 4y = 24

x + y = 6

By **solving** both equations, we **get:**

2y = 8

y = 4

Now, **plugging** y = 4 in other **equation,**

x + 4 = 6

x = 2

Hence, the **circumcenter** of a triangle is (x, y) = (2, 4).

*All images/mathematical drawings were created with GeoGebra.*