The locus of the centroid of the triangle whose vertices are $ (3 \cos \alpha, 3 \sin \alpha),(9 \sin \alpha,-9 \cos \alpha) $ and $ (1,0) $ is a circle of radius $ R $, then the value of $ \frac{9 R^{2}}{2} $ is equal to

Question

The locus of the centroid of the triangle whose vertices are $ (3 \cos \alpha, 3 \sin \alpha),(9 \sin \alpha,-9 \cos \alpha) $ and $ (1,0) $ is a circle of radius $ R $, then the value of $ \frac{9 R^{2}}{2} $ is equal to,

MARKS App · Accepted Answer

Let centroid of ∆ A B C is h , k ⇒ h = 3 cos α + 9 sin α + 1 3 & k = 0 + 3 sin α - 9 cos α 3 ⇒ cos α + 3 sin α = 3 h - 1 & sin α - 3 cos α = 3 k ⇒ cos α + 3 sin α 2 = 3 h - 1 2 & sin α - 3 cos α 2 = 9 k 2 ⇒ 3 h - 1 2 + 9 k 2 = cos α + 3 sin α 2 + sin α - 3 cos α 2 Replace h by x and k by y ⇒ 3 x - 1 2 + 9 y 2 = 10 cos 2 α + sin 2 α ⇒ x - 1 3 2 + y 2 = 10 9 ⇒ R = 10 3 ⇒ 9 R 2 2 = 5

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