JEE Mains · Maths · STD 11 - 4.1 complex nubers
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 \(.............\).
- A \(4\)
- B \(6\)
- C \(2\)
- D \(8\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
For circle : \(\left|z-z_1\right|^2+\left|z-z_2\right|^2=\left|z_1-z_2\right|^2\) \(r =\frac{\left|z_1-z_2\right|}{2}=\frac{|\alpha-\beta|}{2}=\sqrt{\lambda-1}\) \(2 \lambda=|\alpha-\beta|^2\) \(|\alpha-\beta|=2 \sqrt{\lambda-1}\) \(|\alpha-\beta|^2=4 \lambda-4=2 \lambda\)…
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