Question

Find the circumcenter of the triangle whose vertices are (–2, –3), (–1, 0), (7, –6).

Answer 1

The circumference of a triangle is equidistance from the vertices of a triangle.

Let A(-2, -3), B(-1, 0) and C(7, -6) vertices of the given triangle and let P(x,y) be the circumference of this triangle, Then

PA = PB = PC

Now, PA = PB

`=> sqrt((-2-x)^2 + (-3 -y)^2) = sqrt((-1  - x)^2 + (0 - y)^2`

`=> 4 + x^2 + 4x + 9 + y^2 + 6y = 1 + x^2 + 2xz + y^2`

`=> 4 + x^2 + 4x + 9 + y^2 + 6y - 1 - x^2 - 2x - y^2 = 0`

`=> 2x + 6y + 12 = 0`

`=> 2(x + 3y + 6) = 0`

`=> x + 3y + 6 = 0`   ...Equation (1)

And PB = PC

`=> sqrt((-1-x)^2 + (0 - y)^2) = sqrt((7- x)^2 + (-6 - y)^2)`

Squaring both the sides

`=> (-1 - x)^2 + y^2 = (7 - x)^2 + (-6 -y)^2`

`=> 1 + x^2 + 2x + y^2 = (7 - x)^2 + (-6 - y)^2`

`=> 1 + x^2 + 2x + y^2 - 49 - x^2 + 14x - 36 - y^2 - 12y`

`=> 16x - 12y - 84 = 0`

`=> 4(4x - 3y - 21) = 0`

`=> 4x - 3y - 21 = 0`   ...Equation (2)

Adding equation (1) and (2)

`=> x + 3y + 6 + 4x - 3y - 21 = 0` 

`=> x + 3y + 6 + 4x -3y - 21 = 0`

`=> 5x - 15 = 0`

`=> x = 15/5`

`=> x = 3`

Putting the value of x in equation (2), we get

`=> 4 xx 3 - 3y - 21 = 0` 

`=> 12 - 3y - 21 = 0`

`=> -3y - 9 = 0`

`=> y = (-9)/3 = -3`

So the coordinates of the circumcentre P are (3, -3).