Question

Consider a triangle $ABC$ whose one side lies on the line $x+y=4$. If the coordinates of the orthocentre and the centroid are $(0,0)$ and $(2,4)$ respectively, and $R$ is the circumradius, then the value of $3R^2$ is

Answer 1

Let the circumcentre be $O(x,y)$.
Orthocentre is $H(0,0)$ and centroid $G$ is $(2,4)$

$G$ divides $HO$ in the ratio $2:1$.
$\Rightarrow \frac{2x+0}{3}=2\Rightarrow x=3$
$\Rightarrow \frac{2y+0}{3}=4\Rightarrow y=6$
$\therefore$ Coordinates of the circumcentre is $O(3,6).$
We know that the image of the orthocentre about any side lies on the circumcircle.

Finding image of $H(0,0)$ about the line $x+y=4:$
$\frac{x-0}{1}=\frac{y-0}{1}=\frac{-2(0+0-4)}{1^2+1^2}$
Hence, image of $H$ is $(4,4)$.

The distance between $O$ and image of $H$ is the circumradius.
$\therefore$ Circumradius, $R=\sqrt{(4-3{)}^2+(4-6{)}^2}=\sqrt{5}$