The point of concurrency of the perpendicular bisectors of the sides of a triangle is called the circumcenter of the triangle.

The following steps will be useful to find circumcenter of a triangle.

**Step 1 : **

Find the equations of the perpendicular bisectors of any two sides of the triangle.

**Step 2 : **

Solve the two equations found in step 2 for x and y.

The solution (x, y) is the circumcenter of the triangle given.

**Example : **

Find the co ordinates of the circumcenter of a triangle whose vertices are (2, -3), (8, -2) and (8, 6).

**Solution : **

**Let A(2, -3), B(8, -2) and C(8, 6) be the vertices of the triangle. **

**D is the midpoint of AB and E is the midpoint of BC. **

Midpoint of AB is

= [(x_{1} + x_{2})/2, (y_{1} + y_{2})/2]

Substitute (x_{1}, y_{1}) = (2, -3) and (x_{2}, y_{2}) = (8, -2).

= [(2 + 8)/2, (-3 - 2)/2]

= [10/2, -5/2]

= (5, -5/2)

So, the point D is (5, -5/2).

Slope of AB is

= [(y_{2} - y_{1})/(x_{2} - x_{1})]

Substitute (x_{1}, y_{1}) = (2, -3) and (x_{2}, y_{2}) = (8, -2).

= [(-2 - (-3)] / (8 - 2)

= (-2 + 3) / 6

= 1/6

Slope of the perpendicular line to AB is

= -1 / slope of AB

= -1 / (1/6)

= -1 ⋅ (6/1)

= -6

**Equation of the perpendicular bisector to the side AB :**

y = mx + b

Substitute m = -6.

y = -6x + b -----(1)

Substitute the point D(5, -5/2) for (x, y) into the above equation.

-5/2 = -6(5) + b

-2.5 = -30 + b

Add 30 to each side.

27.5 = b

Substitute b = 27.5 in (1).

(1)-----> y = -6x + 27.5

Equation of the perpendicular line through D is

**y = -6x + 27.5 -----(2)**

Midpoint of BC is

= [(x_{1} + x_{2})/2, (y_{1} + y_{2})/2]

Substitute (x_{1}, y_{1}) = (8, -2) and (x_{2}, y_{2}) = (8, 6).

= [(8 + 8)/2, (-2 + 6)/2]

= [16/2, 4/2]

= (8, 2)

So, the point E is (8, 2).

Slope of BC is

= [(y_{2} - y_{1})/(x_{2} - x_{1})]

Substitute (x_{1}, y_{1}) = (8, -2) and (x_{2}, y_{2}) = (8, 6).

= [6 - (-2)] / (8 - 8)

= (6 + 2) / 0

= 8/0

Slope of the perpendicular line to BC is

= -1 / slope of BC

= -1 / (8/0)

= -1 ⋅ (0/8)

= -1 ⋅ 0

= 0

**Equation of the perpendicular bisector to the side BC :**

y = mx + b

Substitute m = 0.

y = b -----(3)

Substitute the point E(8, 2) for (x, y) into the above equation.

2 = b

Substitute b = 2 in (1).

(1)-----> y = 2

Equation of the perpendicular line through D is

**y = 2 -----(4)**

Solving (2) and (4), we get

x = 4.25 and y = 2

Therefore, the circumcenter of the triangle ABC is

(4.25, 2)

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