For $\alpha, \beta, z \in \mathbb{C}$ 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 __________.

Given equation of circle, <br/><br/>$$ \begin{aligned} & \quad|z-\alpha|^2+|z-\beta|^2=2 \lambda \\\\ & \therefore 2 \lambda=|\alpha-\beta|^2 .........(i) \end{aligned} $$ <br/><br/>For circle, <br/><br/>$\left|z-z_1\right|^2+\left|z-z_2\right|^2=\left|z_1-z_2\right|^2$ <br/><br/>$\begin{array}{lll}\text { Radius, } r=\frac{\left|z_1-z_2\right|}{2}=\frac{|\alpha-\beta|}{2}=\sqrt{\lambda-1} \\\\ \Rightarrow |\alpha-\beta|=2 \sqrt{\lambda-1} \\\\ \Rightarrow |\alpha-\beta|^2=4(\lambda-1) \\\\ \Rightarrow 2 \lambda=4(\lambda-1) [\because \text { Using Eq. (i)] } \\\\ \Rightarrow 2 \lambda=4 \\\\ \Rightarrow \lambda=2 \end{array}$ <br/><br/>$\begin{array}{ll}\Rightarrow |\alpha-\beta|^2=4 \\\\ \Rightarrow |\alpha-\beta|=2\end{array}$