Question

Let $z_{1}and{z}_2$ be two non real complex cube roots of unity and ${|z-z_1|}^2+{|z-z_2|}^2=\lambda$ be the equation of a circle with $z_1,z_2$ as ends of a diameter, then the value of $\lambda$ is

• A. $4$
• B. $3$
• C. $2$
• D. $\sqrt{2}$

Answer 1

Answer: (B)

$3$

${|z-z_1|}^2+{|z-z_2|}^2=\lambda$

$z_1$ and $z_2$ are roots of unity.

$\Rightarrow z_1=\frac{-1}{2}+\frac{\sqrt{3}}{2}i$

$z_2=\frac{-1}{2}-\frac{\sqrt{3}}{2}i$

As $z_1,z_2$ are ends of diameter, hence equation of circle is

$\Rightarrow (x+\frac{1}{2})(x+\frac{1}{2})+(y-\frac{\sqrt{3}}{2})(y-\frac{\sqrt{3}}{2})=0$

$\Rightarrow x^2+\frac{1}{4}+x+y^2-\frac{3}{4}=0$

$\therefore x^2+x+y^2-\frac{1}{2}=0$

$2x^2+2x+2y=1$

$\Rightarrow x^2+2x+2y=1$

$\Rightarrow {(x+\frac{1}{2})}^2+{(y-\frac{\sqrt{3}}{2})}^2+{(x+\frac{1}{2})}^2+{(y+\frac{\sqrt{3}}{2})}^2=1+\frac{1}{4}+\frac{1}{4}+\frac{3}{4}+3$

${(x+\frac{1}{2})}^2+{(y-\frac{\sqrt{3}}{2})}^2+{(x+\frac{1}{2})}^2+{(y+\frac{\sqrt{3}}{2})}^2=3$

$\Rightarrow {|z-z_1|}^2+{|z-z_2|}^2=3$

$\therefore \lambda =3$

Hence, the answer is 3.