Mathematics - Complex Number Question with Solution | TestHub

Given,

$\alpha , \beta , z\in \mathrm{ℂ}$ and $\lambda >1$,

And $\sqrt{\lambda -1}$ is the radius of the circle ${\left|z-\alpha \right|}^2+{\left|z-\beta \right|}^2=2\lambda$ 

Now we know that equation of circle is given by,

${\left|z-z_1\right|}^2+{\left|z-z_2\right|}^2={\left|z_1-z_2\right|}^2$, where radius is given by $r=\frac{\left|z_1-z_2\right|}{2}$,

On comparing with ${\left|z-\alpha \right|}^2+{\left|z-\beta \right|}^2=2\lambda$, we get

Radius $\frac{\left|z_1-z_2\right|}{2}=\frac{\left|\alpha -\beta \right|}{2}$ and ${\left|\alpha -\beta \right|}^2=2\lambda$

$\Rightarrow \frac{\left|\alpha -\beta \right|}{2}=\sqrt{\lambda -1} \left(\text{given}\right)$

$\Rightarrow \left|\alpha -\beta \right|=2\sqrt{\lambda -1}$ 

$\Rightarrow {\left|\alpha -\beta \right|}^2=4\left(\lambda -1\right)$

$\Rightarrow 2\lambda =4\left(\lambda -1\right)$

$\Rightarrow \lambda =2$

Hence, ${\left|\alpha -\beta \right|}^2=4\Rightarrow \left|\alpha -\beta \right|=2$