Question

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

Answer 1

Let A(4,6), B(0,4), C(6,2) be the vertices of the given $△$ABC.

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

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

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

$12x-4y=24$

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

Solving equation 1 and 2, we get,

$x=3,y=3$

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

Circumradius = $PA=\sqrt{(4-3{)}^2+(6-3{)}^2}=\sqrt{10}units$