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For $\mathrm{n \in \mathbf{N}}$ , let $\mathrm{\mathrm{S}_{\mathrm{n}}=\left\{z \in \mathrm{C}:|z-3+2 i|=\frac{\mathrm{n}}{4}\right\}}$ and $\mathrm{\mathrm{T}_{\mathrm{n}}=\left\{z \in \mathrm{C}:|z-2+3 i|=\frac{1}{\mathrm{n}}\right\}}$ . Then the number of elements in the set $\mathrm{\left\{\mathrm{n} \in \mathbf{N}: \mathrm{S}_{\mathrm{n}} \cap \mathrm{T}_{\mathrm{n}}=\phi\right\}}$ is :
Option: 1
$0$

Option: 2
$2$

Option: 3
$3$

Option: 4
$4$

Answers (1)

$\mathrm{s_{n}:|z-(3-2 i)|=\frac{n}{4}}$ is a circle centre $\mathrm{c_{1}(3,-2)}$ and radius $\mathrm{\left ( \frac{n}{4} \right )}$ .

$\mathrm{T_{n}:|z-(2-3 i)|=\frac{1}{n}}$ is a circle centre $\mathrm{c_{2}(2,-3)}$ and radius $\mathrm{\left ( \frac{1}{n} \right )}$

Here $\mathrm{S_{n} \cap T_{n}=\phi}$

Both circles do not intersect each other

$\mathrm{\text { case-1 }: C_{1} C_{2}>\frac{n}{4}+\frac{1}{n} }$

$\mathrm{\sqrt{2}>\frac{n}{4}+\frac{1}{n} } \\$

$\mathrm{\text { then } n=1,2,3,4}$

$\mathrm{Case -2 : C_{1} C_{2}<\left|\frac{n}{4}-\frac{1}{n}\right|}$

$\mathrm{\Rightarrow \sqrt{2}<\left|\frac{n^{2}-4}{4 n}\right| \text {. } }$

$\Rightarrow \mathrm{n}$ has infinite solutions for $\mathrm{n \in N}$

Hence correct option is 2

Posted by

Ramraj Saini

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