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B
x2+y2+4x−3=0
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C
x2+y2−4x−3=0
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D
x2+y2−4x+3=0
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Solution
The correct option is Dx2+y2−4x+3=0 Given : z=32+cosθ+isinθ ⇒3z=2+cosθ+isinθ⇒cosθ+isinθ=3z−2 Taking mod on both sides, we get ⇒1=|3−2z||z|⇒|z|=|3−2z|⇒x2+y2=(3−2x)2+4y2⇒3x2+3y2−12x+9=0⇒x2+y2−4x+3=0