Question

Find the circumcentre of the triangle formed by the point $(1,3),(0,2),(-3,1)$

Answer 1

Consider the following question.

Circumcentre of triangle is equidistant from vertices , so

$A(1,3),B(0,-2),C(-3,1)$ are equidistant from $O$

Let the coordinate of $Obe(x,y).$

$AO=BO$

$(x-1){}^{\text{2}}+(y-3){}^{\text{2}}=(x-0){}^{\text{2}}+(y+2){}^{\text{2}}$

$x{}^{\text{2}}+1-2x+y{}^{\text{2}}+9-2y=x{}^{\text{2}}+y{}^{\text{2}}+4-4y$

$10-2x-2y=4-4y$

$2x-2y=6$

$x-y=3.......eq\left(i\right)$

$CO=AO$

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

$x{}^{\text{2}}+1-2x+y{}^{\text{2}}+9-6y=x{}^{\text{2}}+9+6x+y{}^{\text{2}}+1-2y$

$-6y-2x=6x-2y$

$8x+4y=0$

$2x+y=0....eq\left(ii\right)$

By elimination method, we get.

$3x=3$

$x=1$

Put the value of $x$ eq. $\left(i\right)$, we get.

$1-y=3$

$y=-2$

Coordinate of circumcentre $=(1,-2)$

Hence, this is the required answer.