Question

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

• A. 0
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A), (B), (C), (D)

$S_n:|z-(3-2 i)|=\frac{n}{4}$ is a circle center $C _1(3,-2)$ and radius $n / 4$
$T _{ n }:| z -(2-3 i )|=\frac{1}{ n }$ is a circle center $C _2$ $(2,-3)$
and radius $1 / n$
Here $S_n \cap T_n=\phi$
Both circles do not intersect each other
Case-1 : $C _1 C _2> n / 4+1 / n$
$\sqrt{2}>\frac{ n }{4}+\frac{1}{ n }$
then $n =1,2,3,4$
Case-2 : $C _1 C _2<\left|\frac{ n }{4}-\frac{1}{ n }\right|$
$\Rightarrow \sqrt{2} < \left|\frac{n^2-4}{4 n}\right|$
$\Rightarrow n$ has infinite solutions for $n \in N$