Question

If $|z+1|=\sqrt{2}|z-1|$, then locus of z is

• A. straight line
• B. circle
• C. ellipse
• D. parabola

Answer 1

Answer: (B)

circle

we have $|z+1|=\sqrt{2}|z-1|$
squaring $|z+1{|}^2=2|z-1{|}^2$
Let $z=x+iy$
$(x+1{)}^2+y^2=2(x-1{)}^2+2y^2$
$x^2+2x+y^2+1=2x^2-4x+2+2y^2$
$\therefore x^2-6x+y^2+1=0$ is an equation of circle