Q86 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Mathematics
The orthocentre and the centroid of $\triangle A B C$ are $(5,8)$ and $\left(3, \frac{14}{3}\right)$ respectively. The equation of the side $B C$ is $x-y=0$. Given that 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 $\triangle A B C$ is
- A$\sqrt{10}$
- B$2 \sqrt{10}$
- Correct answerC$4 \sqrt{10}$
- D$8 \sqrt{10}$
Solution
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
$$
\begin{array}{rlrl}
& & \left(\frac{2 h+5}{3}, \frac{2 k+8}{3}\right)=\left(3, \frac{14}{3}\right) \\
\Rightarrow \quad & h=2 k=3
\end{array}
$$
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 $=4 \sqrt{10}$.
Hence, the correct answer is option: 'C'.