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 D_{1} be distance between the vertex (x_{1},y_{1}) 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 D_{1}=D_{2 } and D_{2}=D_{3 } , 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-y_{1}) = m (x-x_{1})
- 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
D_{1 }be the distance from the circumcenter to vertex A
D_{2 }be the distance from the circumcenter to vertex B
D_{3} be the distance from the circumcenter to vertex C
Given : (x_{1} , y_{1})=(3,2) ; (x_{2 }, y_{2})=(1,4) and (x_{3} , y_{3})=(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 D_{1}= D_{2} = D_{3 }.
D_{1}= D_{2} 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=1â€¦..(1)
D_{1}= D_{3 }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 =7â€¦..(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)
For more maths concepts, keep visiting BYJUâ€™S and get various maths related videos to understand the concept in an easy and engaging way.
Related Links | |
Circumcenter Formulas | Linear Equations |
Right Angle Triangle | Triangles |