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If $A (\cos α, \sin α), B (\sin α, - \cos α) C (1, 2)$ are vertices of $Δ A B C$ , then as $α$ varies the locus of its centroid is

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$x^{2} + y^{2} - 2 x - 4 y + 3 = 0$

$x^{2} + y^{2} - 2 x - 4 y + 1 = 0$

$3 (x^{2} + y^{2}) - 2 x - 4 y + 1 = 0$

$x^{2} + y^{2} - 2 x + 3 y + 3 = 0$

answer is C.

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Detailed Solution

Centroid $= (h, k)$

$3 h = \cos α + \sin α + 1 \Rightarrow (3 h - 1) = \cos α + \sin α 3 k = \sin α - \cos α + 2 \Rightarrow 3 k - 2 = \sin α - \cos α ∴ {(3 h - 1)}^{2} + {(3 k - 2)}^{2} = 1$

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