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B
circle
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C
ellipse
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D
parabola
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Solution
The correct option is A circle we have |z+1|=√2|z−1| squaring |z+1|2=2|z−1|2 Let z=x+iy (x+1)2+y2=2(x−1)2+2y2 x2+2x+y2+1=2x2−4x+2+2y2 ∴x2−6x+y2+1=0 is an equation of circle