Question

Find the circumcentre of the triangle whose vertices are $(-2,-3),(-1,0),(7,-6)$.

Answer 1

Let $A(-2,-3)$, $B(-1,0)$ and $C(7,-6)$ be the vertices of the triangle. $P(x,y)$ be the required circumcentre.

Also, $P{A}^2=P{B}^2=P{C}^2=Radius$

Distance between two points = $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

So,

$PA=\sqrt{{(x-(-2\left)\right)}^2+{(y-(-1\left)\right)}^2}$

$P{A}^2={(x+2)}^2+{(y+1)}^2$

$P{A}^2=x^2+4+4x+y^2+9+6y$..........(1)

$PB=\sqrt{{(x-(-1\left)\right)}^2+{(y-(0\left)\right)}^2}$

$P{B}^2={(x+1)}^2+y^2$

$P{B}^2=x^2+1+2x+y^2$...........(2)

$PC=\sqrt{{(x-7)}^2+{(y-(-6\left)\right)}^2}$

$P{C}^2={(x-7)}^2+(y+6{)}^2$

$P{C}^2=x^2+49-14x+y^2+36+12y$...........(3)

from (1) and (2) we get,

$\Rightarrow x^2+4+4x+y^2+9+6y=x^2+1+2x+y^2$

$\Rightarrow 4x+6y+13=1+2x$

$\Rightarrow 2x+6y+12=0$

$\Rightarrow x+3y+6=0$

from (1) and (3) we get,

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

$\Rightarrow 1+2x=49-14x+36+12y$

$\Rightarrow 2x-12y+14x=84$

$\Rightarrow 16x-12y=84$

$\Rightarrow 4x-3y=21$

$4(-3y-6)-3y=21$

$\Rightarrow -12y-24-3y=21$

$\Rightarrow 15y=-45$

$y=\frac{-45}{15}=-3$

$x=-6+9=3$

$\therefore verticesofcircumcentre=(3,-3)$