What is the Circumcenter that a Triangle?

The circumcenter that a triangle is defined as the suggest where the perpendicular bisectors the the sides of that particular triangle intersect. In other words, the point of concurrency the the bisector of the political parties of a triangle is referred to as the circumcenter. The is denoted by P(X, Y). The circumcenter is additionally the center of the circumcircle of the triangle and also it deserve to be either inside or outside the triangle.

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Circumcenter Formula


P(X, Y) = <(x1 sin 2A + x2 sin 2B + x3 sin 2C)/ (sin 2A + sin 2B + sin 2C), (y1 sin 2A + y2 sin 2B + y3 sin 2C)/ (sin 2A + sin 2B + sin 2C)>

Here,

A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of the triangle and A, B, C space their particular angles.

Method to calculation the Circumcenter the a Triangle

Steps to uncover the circumcenter that a triangle are:

Calculate the midpoint of offered coordinates, i.e. Midpoints that AB, AC, and BCCalculate the slope of the certain lineBy making use of the midpoint and also the slope, find out the equation that the line (y-y1) = m (x-x1)Find out the equation of the various other line in a comparable mannerSolve two bisector equations by finding the end the intersection pointCalculated intersection allude will it is in the circumcenter the the offered triangle

Finding Circumcenter Using straight Equations

The circumcenter can additionally be calculated by forming linear equations using the distance formula. Let united state take (X, Y) it is in the collaborates of the circumcenter. According to the circumcenter properties, the distance of (X, Y) from each vertex the a triangle would certainly be the same.

Assume the D1 be the distance in between the peak (x1, y1) and the circumcenter (X, Y), climate the formula is provided by,

D1= √<(X−x1)2+(Y−y1)2>D2= √<(X−x2)2+(Y−y2)2>D3= √<(X−x3)2+(Y−y3)2>Learn More: Distance in between Two Points

Now, due to the fact that D1=D2 and also D2=D3, we get

(X−x1)2 + (Y−y1)2 = (X−x2)2 + (Y−y2)2

From this, two linear equations room obtained. By resolving the linear equations using substitution or remove method, the works with of the circumcenter have the right to be obtained.

Properties the Circumcenter

Some the the properties of a triangle’s circumcenter room as follows:

The circumcenter is the centre of the circumcircleAll the vertices the a triangle are equidistant indigenous the circumcenterIn an acute-angled triangle, circumcenter lies within the triangleIn an obtuse-angled triangle, the lies exterior of the triangleCircumcenter lies at the midpoint the the hypotenuse next of a right-angled triangle

How to construct Circumcenter of a Triangle?

The circumcenter of any kind of triangle have the right to be built by drawing the perpendicular bisector of any of the 2 sides of that triangle. The actions to build the circumcenter are:

Step 1: Draw the perpendicular bisector of any two sides of the provided triangle.Step 2: Using a ruler, prolong the perpendicular bisectors until they intersect each other.Step 3: mark the intersecting suggest as p which will be the circumcenter that the triangle. It should be noted that, also the bisector the the third side will also intersect at P.

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Example question Using Circumcenter Formula

Question: Find the works with of the circumcenter the a triangle ABC through the vertices A = (3, 2), B = (1, 4) and also C = (5, 4)?

Solution:

Method 1:

Let, (x, y) be the works with of the circumcenter.

D1 be the street from the circumcenter to vertex A

D2 it is in the street from the circumcenter to vertex B

D3 it is in the street from the circumcenter to vertex C

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

Using distance formula, we get

D1= √<(X−x1)2+(Y−y1)2>D2= √<(X−x2)2+(Y−y2)2>D3= √<(X−x3)2+(Y−y3)2>Since D1= D2 = D3 .

D1= D2 gives,

(x – 3)2 + (y − 2)2 = (x − 1)2 + (y − 4)2

⇒ x2 − 6x + 9 + y2 + 4 − 4y = x2 + 1 – 2x + y2 – 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

⇒ x2 − 6x + 9 + y2 + 4 − 4y = x2 + y2 − 10x – 8y + 25 + 16

⇒ -6x – 4y + 13 = -10x – 8y + 41

⇒ 4x + 4y = 28

Or, x + y = 7 ————–(2)

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

2y = 8

Or, y = 4

Now, substitute y = 4 in equation(1),

⇒ -x + 4 = 1

⇒ -x = 1 – 4

⇒ -x = -3

Or, x = 3

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

Method 2:

Given points are,

A = (3, 2),

B = (1, 4),

C = (5, 4)

To discover out the circumcenter we need to solve any kind of two bisector equations and find the end the intersection points.

So, mid point of ab = <(3 + 1)/2, (2 + 4)/2> = (2, 3)

Slope of abdominal muscle = <(4−2)/(1−3)> = -1

The steep of the bisector is the negative reciprocal of the given slope.

So, the slope of the perpendicular bisector = 1

Equation of abdominal with slope 1 and the collaborates (2, 3) is,

(y – 3) = 1(x – 2)

x – y = -1………………(1)

Similarly, for AC

Mid allude of AC = <(3 + 5)/2, (2 + 4)/2> = (4, 3)

Slope the AC = <(4−2)/(5−3)> = 1

The steep of the bisector is the negative reciprocal the the given slope.

So, the slope of the perpendicular bisector = -1

Equation the AC v slope -1 and also the works with (4, 3) is,

(y – 3) = -1(x – 4)

y – 3 = -x + 4

x + y = 7………………(2)

By resolving equation (1) and (2),

(1) + (2) ⇒ 2x = 6;

Or, x = 3

Substitute the worth of x in to (1)

3 – y = -1

y = 3 + 1 = 4

Thus, the circumcenter is (3, 4).


The circumcenter is the intersection suggest of the perpendicular bisectors of political parties of a triangle. That is the centre of a triangle’s circumcircle.


To uncover the circumcenter of any type of triangle, draw the perpendicular bisectors of the sides and also extend them. The point at i m sorry the perpendicular intersects each various other will be the circumcenter of the triangle.


Yes, every triangle has actually a circumcenter. The circumcenter can be either inside the triangle or outside.

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The circumcenter of an obtuse-angled triangle is outside the triangle. Because that a right-angled triangle, the circumcenter lies top top the hypotenuse.