#### Question

Find the coordinates of the circumcentre of the triangle whose vertices are (3, 0), (-1, -6) and (4, -1). Also, find its circumradius.

#### Solution

The distance *d* between two points `(x_1,y_1)` and `(x_2, y_2)`is given by the formula

`d = sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)`

The circumcentre of a triangle is the point which is equidistant from each of the three vertices of the triangle.

Here the three vertices of the triangle are given to be *A*(3*,*0), *B*(*−*1*,**−*6) and *C*(4*,**−*1)

Let the circumcentre of the triangle be represented by the point *R*(*x, y*).

So we have AR = BR = CR

`AR = sqrt((3 - x)^2 + (-y)^2)`

`BR = sqrt((-1-x)^2 + (-6 -y)^2)`

`CR = sqrt((4 -x)^2 + (-1-y)^2)`

Equating the first pair of these equations we have,

AR= BR

`sqrt((3 - x)^2 + (-y)^2) = sqrt((-1-x)^2 +(-6-y)^2)`

Squaring on both sides of the equation we have,

`sqrt((3 - x)^2 + (-y)^2) = sqrt((-1-x)^2 + (-6-y))`

`9 + x^2 - 6x + y^2 = 1 + x^2 + 2x + 36 + y^2 + 12y`

8x + 12y = -28

2x + 3y = -7

Equating another pair of the equations we have,

AR = CR

`sqrt((3 - x)^2 + (-y)^2) = sqrt((4 - x)^2 + (-1 - y)^2)`

Squaring on both sides of the equation we have,

`(3 - x)^2 + (-y)^2 = (4 - x)^2 + (-1 - y)^2`

`9 + x^2 - 6x + y^2 = 16 + x^2 - 8x + 1 + y^2 + 2y`

2x - 2y = 8

x - y = 4

Now we have two equations for ‘*x*’ and ‘*y*’, which are

2x + 3y = -7

x - y = 4

From the second equation we have y = x - 4. Substituting this value of ‘*y*’ in the first equation we have,

2x + 3(x - 4) = -7

2x + 3x - 12 = -7

5x = 5

x= 1

Therefore the value of ‘*y*’ is,

y = x - 4

= 1 - 4

y = -3

Hence the co-ordinates of the circumcentre of the triangle with the given vertices are (1, -3).

The length of the circumradius can be found out substituting the values of ‘*x*’ and ‘*y*’ in ‘*AR*’

`AR = sqrt((3 - x)^2 + (-y)^2)`

`= sqrt((3 -1)^2 + (3)^2)`

`= sqrt((2)^2 +(3)^2)`

`= sqrt(4 + 9)`

`AR = sqrt13`

Thus the circumradius of the given triangle is `sqrt13` units