The circumcentre of a triangle lies at the origin and its centroid is the midpoint of the line segment…

The mid-point of a line segment joining the points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

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  $G\equiv \left(\frac{a^2+1+2a}{2}, \frac{a^2+1-2a}{2}\right)$

$\Rightarrow G\equiv \left(\frac{{\left(a+1\right)}^2}{2}, \frac{{\left(a-1\right)}^2}{2}\right).$

Also, given circumcentre $C$ is origin $\Rightarrow C\equiv \left(0, 0\right).$

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 $y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right)$

Thus, the orthocentre lies on the line joining the points $\left(0, 0\right)$ and $\left(\frac{{\left(a+1\right)}^2}{2}, \frac{{\left(a-1\right)}^2}{2}\right)$ and its equation, is

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

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

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