For alpha, beta, z in mathbb{C} and lambda | vedclass

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Question

(C) The given equation is $|z - \alpha|^2 + |z - \beta|^2 = 2\lambda$.Using the identity $|z - \alpha|^2 + |z - \beta|^2 = 2|z - \frac{\alpha + \beta}

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The given equation is $|z - \alpha|^2 + |z - \beta|^2 = 2\lambda$.Using the identity $|z - \alpha|^2 + |z - \beta|^2 = 2|z - \frac{\alpha + \beta}{2}|^2 + \frac{1}{2}|\alpha - \beta|^2$,we rewrite the equation as:$2|z - \frac{\alpha + \beta}{2}|^2 + \frac{1}{2}|\alpha - \beta|^2 = 2\lambda$.Dividing by $2$,we get $|z - \frac{\alpha + \beta}{2}|^2 = \lambda - \frac{1}{4}|\alpha - \beta|^2$.This is the equation of a circle $|z - z_0|^2 = R^2$ where $R^2 = \lambda - \frac{1}{4}|\alpha - \beta|^2$.Given the radius $R = \sqrt{\lambda - 1}$,so $R^2 = \lambda - 1$.Equating the two expressions for $R^2$:$\lambda - \frac{1}{4}|\alpha - \beta|^2 = \lambda - 1$.$-\frac{1}{4}|\alpha - \beta|^2 = -1$.$|\alpha - \beta|^2 = 4$.$|\alpha - \beta| = 2$.

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 $.............$.

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