The circumcenter is widely used in geometry. The circumcenter is where all the perpendicular bisectors meet, join, or intersect, and it is mostly used in circles, polygons, rectangles, and triangles.

Let’s learn about the circumcenter of the triangle, its methods, and properties with a lot of examples.

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**What is the Circumcenter of a Triangle?**

The circumcenter of the triangle is obtained by intersecting the perpendicular bisectors of all of the triangle’s sides. This indicates that the triangle’s perpendicular bisectors are parallel (i.e., meeting at one point). Because all triangles are cyclic and may circumscribe a circle, they all have a circumcenter. Perpendicular bisectors of any two sides of a triangle are drawn to construct the circumcenter of any triangle.

The point at which the perpendicular bisectors of the triangle’s sides connect is said to be the circumcenter of the triangle. In simple words, the circumcenter is the point where the bisector of a triangle’s sides intersects. It is symbolized by the letter Q (x, y). The circumcenter is the center of the triangle’s circumcircle, which might be inside or outside the triangle.

**Method to Find Circumcenter of a Triangle**

Here are some steps to solve the circumcenter of the given triangle.

- Calculate the midpoint of given coordinates, such as the AB, AC, and BC midpoints.
- Calculate the slope of the line in question.
- Find the equation of the line (y-y
_{1}) = m (x-x_{1}) using the midpoint and slope - Find the equation for the opposite line in the same way.
- Find the intersection point of two bisector equations to solve them.
- The result will be the circumcenter of the triangle.

**How to Calculate the Circumcenter of a Triangle?**

There are two methods to solve the circumference of the triangle problems. One is as mentioned above and the other method is applied by using the distance formula.

D = √ ((X – x)^{2} + (Y – y)^{2})

**Example 1**

Calculate the circumcenter of the triangle having points (2, 4), (6, 3), and (1, 2).

**Solution**

**Step 1:** Name the points as X, Y, Z respectively.

X = (2, 4)

Y = (6, 3)

Z = (1, 2)

**Step 2:** Now find the mid points of XY and YZ.

Midpoint of XY = ((x_{1} + x_{2}) / 2, (y_{1} + y_{2}) / 2)

= ((2 + 6)/2, (4 + 3)/2)

= (8/2, 7/2)

Midpoint of XY= (4, 7/2)

Similarly,

Midpoint of YZ = ((x_{2} + x_{3}) / 2, (y_{2} + y_{3}) / 2)

= ((6 + 1)/2, (3 + 2)/2)

Midpoint of YZ = (7/2, 5/2)

**Step 3:** Now calculate the slope of XY and YZ.

We know that y_{2 }– y_{1} = m (x_{2} – x_{1}) is the slope equation.

Slope of XY = y_{2 }– y_{1} / x_{2} – x_{1}

= 3 – 4/6 – 2 = -1/4

Perpendicular bisector slope = -1/ slope of line

Perpendicular bisector slope of XY = -1/ (-1/4) = 4

Similarly,

Slope of YZ = y_{3 }– y_{2} / x_{3} – x_{2}

= 2 – 3/ 1 – 6 = -1/-5 = 1/5

Perpendicular bisector slope = -1/ slope of line

Perpendicular bisector slope of YZ = -1/(1/5) = -5

**Step 4:** Now find the equation of line XY.

the general equation of line is,

y = mx + b

put the slope of XY.

y = 4x + b

now put the midpoints of XY in above equation.

7/2 = 4(4) + b

7/2 = 16 + b

7/2 – 16 = b

-25/2 = b

Put the value of b in above equation.

y = 4x – 25/2 … (1)

**Step 5:** Now find the equation of line YZ.

Put the slope of YZ.

y = -5x + b

now put the midpoints of YZ in above equation.

5/2 = -5(7/2) + b

5/2 = -35/2 + b

5/2 + 35/2 = b

40/2 = b

20 = b

Put the value of b in above equation.

y = -5x + 20 … (2)

**Step 6:** Compare the equations.

y = 4x – 25/2 … (1)

y = -5x + 20 … (2)

we can write as,

4x – 25/2 = -5x + 20

4x + 5x = 25/2 + 20

9x = 65/2

x = 65/18

= 3.6111

**Step 7:** Put the value of x in equation 1.

y = 4x – 25/2 … (1)

y = 4(3.6111) – 25/2

= 14.4444 – 12.5

y = 1.9444

Hence, the circumcenter of the triangle is (3.6111, 1.9444).

To reduce the difficulty of such a long calculation simply use a circumcenter calculator to find the accurate result.

**Alternative method.**

We can also find the above problem by using distance formula.

**Step 1:** Find the distance formula of all the points.

D = √ ((X – x)^{2} + (Y – y)^{2})

D_{1} = √ ((X – x_{1})^{2} + (Y – y_{1})^{2})

D_{1} = √ ((X – 2)^{2} + (Y – 4)^{2})

D_{2} = √ ((X – x_{2})^{2} + (Y – y_{2})^{2})

D_{2} = √ ((X – 6)^{2} + (Y – 3)^{2})

D_{3} = √ ((X – x_{3})^{2} + (Y – y_{3})^{2})

D_{3} = √ ((X – 1)^{2} + (Y – 2)^{2})

**Step 2:** Take D_{1} = D_{2}

√ ((X – 2)^{2} + (Y – 4)^{2}) = √ ((X – 6)^{2} + (Y – 3)^{2})

Cancel the square root.

((X – 2)^{2} + (Y – 4)^{2}) = ((X – 6)^{2} + (Y – 3)^{2})

Open the squares.

(X^{2} – 4X + 4) + (Y^{2} -8Y + 16) = (X^{2} -12X + 36) + (Y^{2} -6Y + 9)

X^{2} – 4X + 4 + Y^{2} -8Y + 16 = X^{2} -12X + 36 + Y^{2} -6Y + 9

Separate the constants and variables.

X^{2} – 4X + Y^{2} -8Y – X^{2} + 12X – Y^{2} + 6Y = 36 + 9 – 4 – 16

**8X – 2Y = 25 … (1)**

**Step 3:** Take D_{1} = D_{3}

√ ((X – 2)^{2} + (Y – 4)^{2}) = √ ((X – 1)^{2} + (Y – 2)^{2})

Cancel the square root.

((X – 2)^{2} + (Y – 4)^{2}) = ((X – 1)^{2} + (Y – 2)^{2})

Open the squares.

(X^{2} – 4X + 4) + (Y^{2} -8Y + 16) = (X^{2} -2X + 1) + (Y^{2} – 4Y + 4)

X^{2} – 4X + 4 + Y^{2} -8Y + 16 = X^{2} -2X + 1 + Y^{2} – 4Y + 4

Separate the constants and variables.

X^{2} – 4X + Y^{2} -8Y – X^{2} + 2X – Y^{2} + 4Y = 1 + 4 – 4 -16

-2X -4Y = -15

**2X + 4Y = 15 … (2)**

**Step 4:** Solve the equations.

8X – 2Y = 25 … (1)

2X + 4Y = 15 … (2)

Multiply eq (1) by 2.

2(8X – 2Y) = 2(25)

16X -4Y = 50

Solve the equations.

16X -4Y = 50

2X + 4Y = 15

18X + 0 = 65

X = 65/18

**X = 3.6111**

**Step 5:** Put the value of x in eq (1).

8X – 2Y = 25

8(3.6111) – 2Y = 25

-2Y = 25 – 28.8888

-2Y = -3.8888

Y = -3.8888/-2

**Y = 1.9444**

Hence, the circumcenter of the triangle is (3.6111, 1.9444).

**Summary **

The point at which the perpendicular bisectors of the triangle’s sides connect is said to be the circumcenter of the triangle. There are two methods of solving the circumference of the triangle. One is by finding the midpoints and the slope and the other is by finding the distance formula. These methods are pretty lengthy but not difficult. Once you solve this problem by hand you can easily solve the problems related to this topic.

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