Let A = { z : | z - 2 | 4 } and B = { z : | z - 2 | + | z + 2 | = 5 } . Then the max { | z 1 - z 2 : z 1 A a n d z 2 B } is: [2026]

Question

$Let A = {z \in ℂ : | z - 2 | \leq 4} and$

$B = {z \in ℂ : | z - 2 | + | z + 2 | = 5} .$

Then the max ${| z_{1} - z_{2} | : z_{1} \in A and z_{2} \in B}$ is: [2026]

Smart Achievers · Accepted Answer

(4)

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

$| z - 2 |$ $+$ $| z + 2 |$ $= 5 \Rightarrow \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

$\Rightarrow \frac{4 x^{2}}{25} + \frac{4 y^{2}}{9} = 1$

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

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