# Find the coordinates of the circumcentre of the triangle whose

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 $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

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

$A R=\sqrt{(3-x)^{2}+(-y)^{2}}$

$B R=\sqrt{(-1-x)^{2}+(-6-y)^{2}}$

$C R=\sqrt{(4-x)^{2}+(-1-y)^{2}}$

Equating the first pair of these equations we have,

$A R=B R$

$\sqrt{(3-x)^{2}+(-y)^{2}}=\sqrt{(-1-x)^{2}+(-6-y)^{2}}$

Squaring on both sides of the equation we have,

$(3-x)^{2}+(-y)^{2}=(-1-x)^{2}+(-6-y)^{2}$

$9+x^{2}-6 x+y^{2}=1+x^{2}+2 x+36+y^{2}+12 y$

$8 x+12 y=-28$

$2 x+3 y=-7$

Equating another pair of the equations we have,

$A R=C R$

$\sqrt{(3-x)^{2}+(-y)^{2}}=\sqrt{(4-x)^{2}+(-1-y)^{2}}$

Squaring on both sides of the equation we have,

$(3-x)^{2}+(-y)^{2}=(4-x)^{2}+(-1-y)^{2}$

$9+x^{2}-6 x+y^{2}=16+x^{2}-8 x+1+y^{2}+2 y$

$2 x-2 y=8$

$x-y=4$

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

$2 x+3 y=-7$

$x-y=4$

From the second equation we have. Substituting this value of ‘y’ in the first equation we have,

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

$2 x+3 x-12=-7$

$5 x=5$

$x=1$

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.

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

$A R=\sqrt{(3-x)^{2}+(-y)^{2}}$

$=\sqrt{(3-1)^{2}+(3)^{2}}$

$=\sqrt{(2)^{2}+(3)^{2}}$

$=\sqrt{4+9}$

$A R=\sqrt{13}$

Thus the circumradius of the given triangle is $\sqrt{13}$ units