1. Let the circle be given by $|z-\alpha|^2+|z-\beta|^2=2\lambda.$ We are told that its radius is $\sqrt{\lambda-1}.$ We need to find $|\alpha-\beta|$.

2. Use the identity for midpoint form.

Let $m=\frac{\alpha+\beta}{2}.$ Then write $z-\alpha=(z-m)-\frac{\alpha-\beta}{2}, \qquad z-\beta=(z-m)+\frac{\alpha-\beta}{2}.$

So, \begin{align*} |z-\alpha|^2+|z-\beta|^2 &=\left|(z-m)-\frac{\alpha-\beta}{2}\right|^2+\left|(z-m)+\frac{\alpha-\beta}{2}\right|^2. \end{align*}

Now use $|u-v|^2+|u+v|^2=2|u|^2+2|v|^2.$

Taking $u=z-m, \qquad v=\frac{\alpha-\beta}{2},$ we get \begin{align*} |z-\alpha|^2+|z-\beta|^2 &=2\left|z-\frac{\alpha+\beta}{2}\right|^2+2\left|\frac{\alpha-\beta}{2}\right|^2 \ &=2\left|z-\frac{\alpha+\beta}{2}\right|^2+\frac{|\alpha-\beta|^2}{2}. \end{align*}

3. Substitute into the given equation: $2\left|z-\frac{\alpha+\beta}{2}\right|^2+\frac{|\alpha-\beta|^2}{2}=2\lambda.$ Divide by $2$: $\left|z-\frac{\alpha+\beta}{2}\right|^2=\lambda-\frac{|\alpha-\beta|^2}{4}.$

Hence the radius of the circle is $r=\sqrt{\lambda-\frac{|\alpha-\beta|^2}{4}}.$

4. Given that the radius is also $\sqrt{\lambda-1},$ so $\lambda-\frac{|\alpha-\beta|^2}{4}=\lambda-1.$

Therefore, $\frac{|\alpha-\beta|^2}{4}=1$ which gives $|\alpha-\beta|^2=4.$ Thus, $|\alpha-\beta|=2.$

5. Final answer: $\boxed{2}$

Comparison with stored correct answer: the stored correct answer is $2$, which matches our result.