JEE MAIN - Mathematics (2022 - 25th July Morning Shift - No. 3)
For $$\mathrm{n} \in \mathbf{N}$$, let $$\mathrm{S}_{\mathrm{n}}=\left\{z \in \mathbf{C}:|z-3+2 i|=\frac{\mathrm{n}}{4}\right\}$$ and $$\mathrm{T}_{\mathrm{n}}=\left\{z \in \mathbf{C}:|z-2+3 i|=\frac{1}{\mathrm{n}}\right\}$$. Then the number of elements in the set $$\left\{n \in \mathbf{N}: S_{n} \cap T_{n}=\phi\right\}$$ is :
0
2
3
4
Explanation
$S_{n}=\left\{z \in C:|z-3+2 i|=\frac{n}{4}\right\}$ represents a circle with centre $C_{1}(3,-2)$ and radius $r_{1}=\frac{n}{4}$
Similarly $T_{n}$ represents circle with centre $C_{2}(2,-3)$ and radius $r_{2}=\frac{1}{n}$
$\text { As } S_{n} \cap T_{n}=\phi$
$$ \begin{array}{llr} C_{1} C_{2}>r_{1}+r_{2}& \text { OR } & C_{1} C_{2}<\left|r_{1}-r_{2}\right| \\\\ \sqrt{2}>\frac{n}{4}+\frac{1}{n} & \text { OR } & \sqrt{2}<\left|\frac{n}{4}-\frac{1}{n}\right| \\\\ n=1,2,3,4&& n \text { may take infinite values } \end{array} $$
Similarly $T_{n}$ represents circle with centre $C_{2}(2,-3)$ and radius $r_{2}=\frac{1}{n}$
$\text { As } S_{n} \cap T_{n}=\phi$
$$ \begin{array}{llr} C_{1} C_{2}>r_{1}+r_{2}& \text { OR } & C_{1} C_{2}<\left|r_{1}-r_{2}\right| \\\\ \sqrt{2}>\frac{n}{4}+\frac{1}{n} & \text { OR } & \sqrt{2}<\left|\frac{n}{4}-\frac{1}{n}\right| \\\\ n=1,2,3,4&& n \text { may take infinite values } \end{array} $$
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