PointHard

Question

If A(cosα, sinα), B (sinα, − cosα), C (1, 2) are the vertices of a ᐃABC, then as α varies, the locus of its centroid is -

Options

A.x2 + y2 − 2x – 4y + 3 = 0
B.x2 + y2 − 2x – 4y + 1 = 0
C.3(x2 + y2 ) − 2x − 4y + 1 = 0
D.None of these

Solution

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