For , ,z C and 1, if -1 is the radius of the circle |z- |^2+|z- |^2=2 , then | - | is equ…

Question

For $\alpha ,\beta ,z\in C$ and $\lambda >1$, if $\sqrt{\lambda -1}$ is the radius of the circle $|z-\alpha {|}^{2}+|z-\beta {|}^{2}=2\lambda $, then $|\alpha -\beta |$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 2

Accepted Answer

Given equation of circle,

$\begin{gathered} \, |z-\alpha {|}^{2}+|z-\beta {|}^{2}=2\lambda \ 	herefore 2\lambda =|\alpha -\beta {|}^{2}.........(i) \end{gathered}$

For circle,

${|z-{z}_{1}|}^{2}+{|z-{z}_{2}|}^{2}={|{z}_{1}-{z}_{2}|}^{2}$

$\begin{gathered} r=\frac{|{z}_{1}-{z}_{2}|}{2}=\frac{|\alpha -\beta |}{2}=\sqrt{\lambda -1} \ \Rightarrow |\alpha -\beta |=2\sqrt{\lambda -1} \ \Rightarrow |\alpha -\beta {|}^{2}=4(\lambda -1) \ \Rightarrow 2\lambda =4(\lambda -1)[\because \ \Rightarrow 2\lambda =4 \ \Rightarrow \lambda =2 \end{gathered}$

$\begin{gathered} \Rightarrow |\alpha -\beta {|}^{2}=4 \ \Rightarrow |\alpha -\beta |=2 \end{gathered}$

Solutions · 1