Question

The orthocentre and the centroid of $\Delta ABC$ are $(5,8)$ and $\left(3,\frac{14}{3}\right)$ respectively. The equation of the side $BC$ is $x-y=0$. Given the image of the orthocentre of a triangle with respect to any side lies on the circumcircle of that triangle, then the diameter of the circumcircle of $\Delta ABC$ is

• A. $\sqrt{10}$
• B. $2\sqrt{10}$
• C. $4\sqrt{10}$
• D. $8\sqrt{10}$

Answer 1

Answer: (C)

The centroid on a non-equilateral triangle divides the line joining orthocentre and circumcentre in $2:1$, so let the coordinates of circumcentre is $(h,k)$, then
$\left(\frac{2h+5}{3},\frac{2k+8}{3}\right)=\left(3,\frac{14}{3}\right)$
$\Rightarrow h=2,k=3$
And image of orthocentre $(5,8)$ with respect to side $BC,x-y=0$ is $(8,5)$
So, the radius of circumcircle of $△ABC$ is
$\sqrt{(8-2{)}^2+(5-3{)}^2}=\sqrt{40}$
Then diameter $=4\sqrt{10}$