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

Given H = (5, 8)  and  G(3, 14/3)

Let S (x, y) be the circumcentre

G divides HS in the ratio 2 : 1

$\begin{array}{l}G=\frac{2s+H}{3}\Rightarrow S=\frac{3G-H}{2}\\ \therefore S=\frac{1}{2}\left(3\left(3,\frac{14}{3}\right)-\left(5,8\right)\right)\\ =\frac{1}{2}\left(9-5,14-8\right)\Rightarrow \frac{1}{2}\left(4,6\right)=\left(2,3\right)\\ \therefore S=\left(2,3\right)\end{array}$

Let  P(h  k) be the Image of H (5,  8) w.r.to x – y = 0

$\begin{array}{l}\frac{h-5}{1}=\frac{k-8}{-1}=\frac{-2\left(5-8\right)}{1+1}\\ \Rightarrow \frac{h-5}{1}=\frac{k-8}{-1}=3\\ h=3+5\Rightarrow 8,k=-3+8\Rightarrow 5\\ \therefore P\left(8,5\right)\\ Diameter=2SP=2\sqrt{{\left(8-2\right)}^2+{\left(5-3\right)}^2}\Rightarrow 2\sqrt{36+4}\Rightarrow 2\sqrt{40}\Rightarrow 4\sqrt{10}\end{array}$