# Circumcenter of a Triangle

The circumcenter of a triangle is the point where the perpendicular bisector of the sides a triangle intersects. In other words, the point of concurrency of the bisector of the sides of a triangle is called the circumcenter and it is denoted by P(X,Y)

## Properties of circumcenter of a triangle

Some of the properties are as follows:

• Circumcenter is the center of circumcircle
• All the vertices of a triangle are equidistant from the circumcenter.
• In acute angled triangle, circumcenter lies inside the triangle.
• In obtuse angled triangle, it lies outside of the triangle
• Circumcenter lies at the mid point of the hypotenuse side of a right angled triangle

## Circumcenter Formula

Let us take (X,Y) be the coordinates of the circumcenter. According to the properties, the distance from each vertex of a triangle would be the same.

Assume that D1 be distance between the vertex (x1,y1) and the circumcenter(X,Y), then the formula is given by

$D_{1}=\sqrt{(X-x_{1})^{2}+(Y-y_{1})^{2}}$ $D_{2}=\sqrt{(X-x_{2})^{2}+(Y-y_{2})^{2}}$ $D_{3}=\sqrt{(X-x_{3})^{2}+(Y-y_{3})^{2}}$

Since D1=D2 and D2=D3 , we get

$(X-x_{1})^{2}+(Y-y_{1})^{2}=(X-x_{2})^{2}+(Y-y_{2})^{2}$

From this,we get two linear equations and solve them using substitution or elimination method for coordinates of the circumcenter.

### Alternate Method to Find Circumcenter of a Triangle

• Find the midpoint of given coordinates like AB, BC and AC
• Find the slope of the required line
• Using the slope and the midpoint, find out the equation of line (y-y1) = m (x-x1)
• Find out the equation of the other line in the same manner
• Find out the intersection point by solving the two bisector equation.
• The calculated intersection point will be the circumcenter of the given triangle.

## Worked Example

### Question:

Find the coordinates of circumcenter of a triangle ABC with the vertices A= (3,2), B= (1,4) and C= (5,4)?

### Solution:

Let, (x,y) be the coordinates of the circumcenter

D1 be the distance from the circumcenter to vertex A

D2 be the distance from the circumcenter to vertex B

D3 be the distance from the circumcenter to vertex C

Given : (x1 , y1)=(3,2) ; (x2 , y2)=(1,4) and (x3 , y3)=(5,4)

Using distance form, we get

$D_{1}=\sqrt{(x-x_{1})^{2}+(y-y_{1})^{2}}=\sqrt{(x-3)^{2}+(y-2)^{2}}$ $D_{2}=\sqrt{(x-x_{2})^{2}+(y-y_{2})^{2}}=\sqrt{(x-1)^{2}+(y-4)^{2}}$ $D_{3}=\sqrt{(x-x_{3})^{2}+(y-y_{3})^{2}}=\sqrt{(x-5)^{2}+(y-4)^{2}}$

Since D1= D2 = D3 .

D1= D2 gives,

$(x – 3)^{2} + (y-2)^{2} = (x -1)^{2} + (y-4)^{2}$ $x^{2}-6x+9+y^{2}+4-4y= x^{2} + 1+ 2x + y^{2} – 8y +16$

-6x-4y+13 =-2x-8y+17

-4x+4y=4

-x+y=1…..(1)

D1= D3 gives,

$(x – 3)^{2}+ (y-2)^{2} = (x-5) ^{2} + ( y – 4 )^{2}$ $x^{2}-6x+9+y^{2}+4-4y = x^{2} + y^{2}-10x – 8y + 25+16$

-6x-4y+13=-10x-8y+41

4x+4y=28

x + y =7…..(2)

By solving equation (1) and (2), we get

2y=8

y=4

Substitute y=4 in equation(1),

-x+4=1

-x=1-4

-x=-3

x=3

Therefore, the circumcenter of a triangle is (x,y)=(3,4)

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