Question

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

• A. 0
• B. 2
• C. 3
• D. 4

Answer 1

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

$S_n:|z-(3-2i)|=\frac{n}{4}$ is a circle center $C_1(3,-2)$ and radius $n/4$
$T_n:|z-(2-3i)|=\frac{1}{n}$ is a circle center $C_2$ $(2,-3)$
and radius $1/n$
Here $S_n\cap T_n=\varphi$
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 : $C_1{C}_2<\left|\frac{n}{4}-\frac{1}{n}\right|$
$\Rightarrow \sqrt{2}<\left|\frac{n^2-4}{4n}\right|$
$\Rightarrow n$ has infinite solutions for $n\in N$