A triangle is drawn such that its circumcentre lies at the origin and the centroid is the mid-point of the line joining the points $(a^{2} + 1, a^{2} + 1)$ and $(2 a, - 2 a)$ . Find the locus of its orthocentre.

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Solution

we know that
$- - \to S H = 3 - - \to S G$
let $H (x, y)$ be orthocentre
$\Rightarrow (x, y) = 3 (\frac{a^{2} + 1 + 2 a}{2}, \frac{a^{2} + 1 - 2 a}{2})$
$\Rightarrow {(a - 1)}^{2} = \frac{2 y}{3}; {(a + 1)}^{2} = \frac{2 x}{3}$
$\Rightarrow 4 a = 2 (\frac{x - y}{3}) \Rightarrow x - y = 6 a \Rightarrow \frac{2 x}{3} = {(\frac{x - y}{6} + 1)}^{2}$
$\Rightarrow {(x - y + 6)}^{2} = 24 x \Rightarrow x^{2} + y^{2} + 36 - 2 x y - 12 y + 12 x = 24 x$
$x^{2} + y^{2} - 2 x y - 12 x - 12 y + 36 = 0$

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