Jee Mains
Let $\mathrm{A}=\{\mathrm{z} \in \mathrm{C}:|\mathrm{z}-2| \leq 4\}$ and $\mathrm{B}=\{\mathrm{z} \in \mathrm{C}:|\mathrm{z}-2|+|\mathrm{z}+2|=5\}$. Then the max $\left\{\left|\mathrm{z}_{1}-\mathrm{z}_{2}\right|: \mathrm{z}_{1} \in \mathrm{~A}\right.$ and $\left.\mathrm{z}_{2} \in \mathrm{~B}\right\}$ is
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Solution
Set A: $|\mathrm{z}-2| \leq 4 \Rightarrow \mathrm{Z}_{1}$ represents interior and
circumference of circle whose
Centre $(2,0)$ and radius $\mathrm{r}=4$
Set B: $|\mathrm{z}-2|+|\mathrm{z}+2|=5$\[ \mathrm{PS}_{1}+\mathrm{PS}_{2}=\ell_{\text {major axis }}=2 \mathrm{a}=5 \]$\mathrm{Z}_{2}$ represents Ellipse with centre origin and major axis as x axis
$\left|\mathrm{z}_{1}-\mathrm{z}_{2}\right|_{\text {max }}=6-\left(-\frac{5}{2}\right)=\frac{17}{2}=$ shortest distance between ellipse and circle lies along x-axis
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