JEE MAIN - Mathematics (2023 - 6th April Evening Shift - No. 18)
For $$\alpha, \beta, z \in \mathbb{C}$$ 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 __________.
Answer
2
Explanation
Given equation of circle,
$$ \begin{aligned} & \quad|z-\alpha|^2+|z-\beta|^2=2 \lambda \\\\ & \therefore 2 \lambda=|\alpha-\beta|^2 .........(i) \end{aligned} $$
For circle,
$$ \left|z-z_1\right|^2+\left|z-z_2\right|^2=\left|z_1-z_2\right|^2 $$
$\begin{array}{lll}\text { Radius, } r=\frac{\left|z_1-z_2\right|}{2}=\frac{|\alpha-\beta|}{2}=\sqrt{\lambda-1} \\\\ \Rightarrow |\alpha-\beta|=2 \sqrt{\lambda-1} \\\\ \Rightarrow |\alpha-\beta|^2=4(\lambda-1) \\\\ \Rightarrow 2 \lambda=4(\lambda-1) [\because \text { Using Eq. (i)] } \\\\ \Rightarrow 2 \lambda=4 \\\\ \Rightarrow \lambda=2 \end{array}$
$\begin{array}{ll}\Rightarrow |\alpha-\beta|^2=4 \\\\ \Rightarrow |\alpha-\beta|=2\end{array}$
$$ \begin{aligned} & \quad|z-\alpha|^2+|z-\beta|^2=2 \lambda \\\\ & \therefore 2 \lambda=|\alpha-\beta|^2 .........(i) \end{aligned} $$
For circle,
$$ \left|z-z_1\right|^2+\left|z-z_2\right|^2=\left|z_1-z_2\right|^2 $$
$\begin{array}{lll}\text { Radius, } r=\frac{\left|z_1-z_2\right|}{2}=\frac{|\alpha-\beta|}{2}=\sqrt{\lambda-1} \\\\ \Rightarrow |\alpha-\beta|=2 \sqrt{\lambda-1} \\\\ \Rightarrow |\alpha-\beta|^2=4(\lambda-1) \\\\ \Rightarrow 2 \lambda=4(\lambda-1) [\because \text { Using Eq. (i)] } \\\\ \Rightarrow 2 \lambda=4 \\\\ \Rightarrow \lambda=2 \end{array}$
$\begin{array}{ll}\Rightarrow |\alpha-\beta|^2=4 \\\\ \Rightarrow |\alpha-\beta|=2\end{array}$
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