Question

Find the coordinates of the circumcentre of the triangle whose vertices are $(8,6),(8,-2)$ and $(2,-2)$. Also, find its circum-radius.

Answer 1

Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let A(8,6), B(8,-2) and C(2,-2) be the vertices of the given triangle and let P(x,y) be the circumcentre of this triangle. Then

$PA=PB=PC$

$\Rightarrow {PA}^2={PB}^2={PC}^2$

$Now,{PA}^2={PB}^2$

$\Rightarrow {(x-8)}^2+{(y-6)}^2={(x-8)}^2+{(y+2)}^2$

$\Rightarrow x^2+y^2-16x-12y+100=x^2+y^2-16x+4y+68$

$\Rightarrow 16y=32$

$\Rightarrow y=2$

$and,{PB}^2={PC}^2$

$\Rightarrow {(x-8)}^2+{(y+2)}^2={(x-2)}^2+{(y+2)}^2$

$\Rightarrow x^2+y^2-16x+4y+68=x^2+y^2-4x+4y+8$

$\Rightarrow 12x=60$

$\Rightarrow x=5$

So, the coordinates of the circumcentre P are (5,2)

Also, Circum-radius=PA=PB=PC=$\sqrt{{(5-8)}^2+{(2-6)}^2}=5$