For α, β, z ∈ ℂ and λ > 1 , if λ - 1 is the radius of the circle |z...

Given,

$\alpha, \beta, z \in \mathbb{C}$ 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 $r = \dfrac{\left|{z}_{1} - {z}_{2}\right|}{2}$,

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

Radius $\dfrac{\left|{z}_{1} - {z}_{2}\right|}{2} = \dfrac{\left|\alpha - \beta\right|}{2}$ and ${\left|\alpha - \beta\right|}^{2} = 2\lambda$

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

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

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

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

$⇒ \lambda = 2$

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