Question

Locus of z such that $|z-a{|}^2+|z-b{|}^2=\lambda$ real is?

• A. Circle
• B. Parabola
• C. Ellipse
• D. Pair of straight

Answer 1

Answer: (A)

Circle

$|z-a{|}^2+|z-b{|}^2=\lambda$

If $z=lx+iy$

$|z|=\sqrt{x^2+y^2}$

$(\sqrt{(x-a{)}^2+y^2}{)}^2+(\sqrt{(x-6{)}^2+y^2}{)}^2=\lambda$

$(x-a{)}^2+(x-b{)}^2+y^2+y^2=\lambda$

coeff of $x^2$ = coeff of $y^2$

coeff of $xy=0$

this is an equation of circle.