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

• Correct answer
  A

  $\frac{17}{2}$
• B

  8
• C

  9
• D

  $\frac{15}{2}$

Solution

$|z - 2| \leq 4 \Rightarrow (x - 2)^2 + y^2 \leq 16$
$|z - 2| + |z + 2| = 5 \Rightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$
$\Rightarrow \dfrac{4x^2}{25} + \dfrac{4y^2}{9} = 1$

Maximum value of $|z_1 - z_2| = 6 + \dfrac{5}{2} = \dfrac{17}{2}$

Hence, the correct answer is option: 'A'.