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QJEEM23Ma4d82: JEE Main 2023 Mathematics - Class 11 JEEM23Ma4d82 Solution: For and , if is the radius of the circle , then is equal to…

For α,β,z∈C\alpha, \beta, z \in \mathbb{C} and \lambda > 1 , if λ−1\sqrt{\lambda-1} is the radius of the circle ∣z−α∣2+∣z−β∣2=2λ|z-\alpha|^{2}+|z-\beta|^{2}=2 \lambda , then ∣α−β∣|\alpha-\beta| is equal to __________.

Step-by-Step Verified Solution

Given equation of circle,

amp;∣z−α∣2+∣z−β∣2=2λamp;∴2λ=∣α−β∣2.........(i)\begin{aligned}& \quad|z-\alpha|^2+|z-\beta|^2=2 \lambda \\\\& \therefore 2 \lambda=|\alpha-\beta|^2 .........(i)\end{aligned}

For circle,

∣z−z1∣2+∣z−z2∣2=∣z1−z2∣2\left|z-z_1\right|^2+\left|z-z_2\right|^2=\left|z_1-z_2\right|^2

 Radius, r=∣z1−z2∣2=∣α−β∣2=λ−1⇒∣α−β∣=2λ−1⇒∣α−β∣2=4(λ−1)⇒2λ=4(λ−1)[∵ Using Eq. (i)] ⇒2λ=4⇒λ=2\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}

⇒∣α−β∣2=4⇒∣α−β∣=2\begin{array}{ll}\Rightarrow |\alpha-\beta|^2=4 \\\\ \Rightarrow |\alpha-\beta|=2\end{array}

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