The circumcentre of a triangle lies at the origin and its centroid is the midpoint of the line segment joining the points (a^2+1,a^2+1) and (2a, -…

The mid-point of a line segment joining the points $({x}_{1},{y}_{1})$(x1​,y1​) and $({x}_{2},{y}_{2})$(x2​,y2​) is $(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2})$(2x1​+x2​​,2y1​+y2​​)

Given, the centroid $G$G is the mid-point of the line segment joining the points $({a}^{2}+1,{a}^{2}+1)$(a2+1,a2+1) and $(2a,-2a)$(2a,−2a), thus $G\equiv (\frac{{a}^{2}+1+2a}{2},\frac{{a}^{2}+1-2a}{2})$G≡(2a2+1+2a​,2a2+1−2a​)

$$\Rightarrow G\equiv (\frac{{(a+1)}^{2}}{2},\frac{{(a-1)}^{2}}{2}).$$⇒G≡(2(a+1)2​,2(a−1)2​).

Also, given circumcentre $C$C is origin $\Rightarrow C\equiv (0,0).$⇒C≡(0,0).

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

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

Again, we know that the equation of a line passing through the points $({x}_{1},{y}_{1})$(x1​,y1​) and $({x}_{2},{y}_{2})$(x2​,y2​) is $y-{y}_{1}=(\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}})(x-{x}_{1})$y−y1​=(x2​−x1​y2​−y1​​)(x−x1​)

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

$$y-0=\frac{(\frac{{(a-1)}^{2}}{2}-0)}{(\frac{{(a+1)}^{2}}{2}-0)}(x-0)$$y−0=(2(a+1)2​−0)(2(a−1)2​−0)​(x−0)

$$\Rightarrow y=\frac{{(a-1)}^{2}}{{(a+1)}^{2}}(x)$$⇒y=(a+1)2(a−1)2​(x)

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