Question

The orthocentre and the centroid of $\Delta A B C$ are $(5,8)$ and $\left(3, \frac{14}{3}\right)$ respectively. The equation of the side $B C$ 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 $\Delta A B C$ 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
$\left(\frac{2 h+5}{3}, \frac{2 k+8}{3}\right)=\left(3, \frac{14}{3}\right)$
$\Rightarrow h=2,\, k=3$
And image of orthocentre $(5,8)$ with respect to side $B C, x-y=0$ is $(8,5)$
So, the radius of circumcircle of $\triangle A B C$ is
$\sqrt{(8-2)^{2}+(5-3)^{2}}=\sqrt{40}$
Then diameter $=4 \sqrt{10}$