If x=2+3cosθ and y=1−3sinθ represent a circle then the centre and radius is?
A
(2,1),9
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B
(2,1),3
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C
(1,2),13
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D
(−2,−1),3
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Solution
The correct option is B(2,1),3 We have, x=2+3cosθ⇒3cosθ⇒3cosθ=x−2 ..(i) y=1−3sinθ⇒3sinθ=−y+1 ..(ii) Squaring adding (i) & (ii), we get (x−2)2(−(y−1))2=32(cos2θ+sin2θ) ⇒(x−2)2+(−(y−1))2=32 ∴ Radius =3, centre =(2,1).