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JEE MAIN 2023
08-04-2023 S2
Question
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 $\_\_\_\_$ . [JEE-Main 08-04-2023_(S2)_CN_M] Toug\#
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