Question

Find the coordinates of the circumcentre of a triangle whose vertices are $A(4,6),B(0,4)$ and $C(6,2).$

Answer 1

Let P(x,y) be the circumcentre of $△ABC$. Then,

$PA=PB=PC$

$P{A}^2=P{B}^2=P{C}^2$

Now, $P{A}^2=P{B}^2$

$(x-4{)}^2+(y-6{)}^2=(x-0{)}^2+(y-4{)}^2$

$x^2+y^2-8x-12y+52=x^2+y^2-8y+16$

$8x+4y=36$

$2x+y=9$ .......(1)

Again, $P{B}^2=P{C}^2$

$(x-0{)}^2+(y-4{)}^2=(x-6{)}^2+(y-2{)}^2$

$x^2+y^2-8y+16=x^2+y^2-12x-4y+40$

$3x-y=6$ ........(2)

Solving equation 1 and 2, we get,

$x=3$ and $y=3$

Therefore, coordinates of circumcentre are $P(3,3)$.