The circumcentre of a triangle lies at the origin and its centroid...

The mid-point of a line segment joining the points $\left({x}_{1}, {y}_{1}\right)$ and \left({x}_{2}, {y}_{2}\right)$is \left(\dfrac{{x}_{1} + {x}_{2}}{2}, \dfrac{{y}_{1} + {y}_{2}}{2}\right)$

Given, the centroid $G$ is the mid-point of the line segment joining the points \left({a}^{2} + 1, {a}^{2} + 1\right)$and \left(2a, - 2a\right)$, thus $G ≡ \left(\dfrac{{a}^{2} + 1 + 2a}{2}, \dfrac{{a}^{2} + 1 - 2a}{2}\right)$

$⇒ G ≡ \left(\dfrac{{\left(a + 1\right)}^{2}}{2}, \dfrac{{\left(a - 1\right)}^{2}}{2}\right) .$

Also, given circumcentre $C$ is origin $⇒ C ≡ \left(0, 0\right) .$

We know that, for a triangle circumcentre, orthocentre and centroid are collinear.

Thus, $C, G$ and orthocentre $H$ are collinear.

Again, we know that the equation of a line passing through the points $\left({x}_{1}, {y}_{1}\right)$ and \left({x}_{2}, {y}_{2}\right)$is$y - {y}_{1} = \left(\dfrac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}\right)\left(x - {x}_{1}\right)$</p><p>Thus, the orthocentre lies on the line joining the points \left(0, 0\right)$ and \left(\dfrac{{\left(a + 1\right)}^{2}}{2}, \dfrac{{\left(a - 1\right)}^{2}}{2}\right) and its equation, is

$y - 0 = \dfrac{\left(\dfrac{{\left(a - 1\right)}^{2}}{2} - 0\right)}{\left(\dfrac{{\left(a + 1\right)}^{2}}{2} - 0\right)}\left(x - 0\right)$

$⇒ y = \dfrac{{\left(a - 1\right)}^{2}}{{\left(a + 1\right)}^{2}}\left(x\right)$

$⇒ {\left(a - 1\right)}^{2}x - {\left(a + 1\right)}^{2}y = 0 .$