Let$A = \{ z \in \mathbb{C}:|z - 2| \leq 4\}$ and$B = \{ z \in \mathbb{C}:|z - 2| + |z + 2| = 5\}$.T

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Let$A = \{ z \in \mathbb{C}:|z - 2| \leq 4\}$ and$B = \{ z \in \mathbb{C}:|z - 2| + |z + 2| = 5\}$.Then the max $\left\{ \left| z_{1} - z_{2} ight|:z_{1} \in 	ext{ }A ight.\ $ and $\left. \ z_{2} \in 	ext{ }B ight\}$ is

Accepted Answer

$|z - 2| \leq 4 \Rightarrow (x - 2) + y^{2} \leq 16$

$$\begin{matrix} & \ |z - 2| + |z + 2| = 5 \Rightarrow \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \\ & \ \Rightarrow \frac{4x^{2}}{25} + \frac{4y^{2}}{9} = 1 \end{matrix}$$

Maximum value of $\left| z_{1} - z_{2} \right| = 6 + \frac{5}{2} = \frac{17}{2}$

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