Find the coordinates of the circumcentre of the triangle whose

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