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 ______.

Given $$\alpha, \beta, z \in \mathbb{C}$$, $$\lambda > 1$$, and $$\sqrt{\lambda - 1}$$ is the radius of the circle $$|z - \alpha|^2 + |z - \beta|^2 = 2\lambda$$.

Expanding the left side:

$$|z|^2 - 2\text{Re}(z\bar{\alpha}) + |\alpha|^2 + |z|^2 - 2\text{Re}(z\bar{\beta}) + |\beta|^2 = 2\lambda$$

$$2|z|^2 - 2\text{Re}\left(z(\bar{\alpha} + \bar{\beta})\right) + |\alpha|^2 + |\beta|^2 = 2\lambda$$

Completing the square by writing $$z = w + \frac{\alpha + \beta}{2}$$:

$$\left|z - \frac{\alpha + \beta}{2}\right|^2 = \lambda - \frac{|\alpha - \beta|^2}{4}$$

This is a circle with center $$\frac{\alpha + \beta}{2}$$ and radius:

$$r = \sqrt{\lambda - \frac{|\alpha - \beta|^2}{4}}$$

Given that the radius equals $$\sqrt{\lambda - 1}$$:

$$\sqrt{\lambda - \frac{|\alpha - \beta|^2}{4}} = \sqrt{\lambda - 1}$$

Squaring both sides:

$$\lambda - \frac{|\alpha - \beta|^2}{4} = \lambda - 1$$

$$\frac{|\alpha - \beta|^2}{4} = 1$$

$$|\alpha - \beta| = 2$$

The answer is $$2$$.