JEE Mains · Maths · STD 11 - 4.1 complex nubers
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 \(\left\{ n \in N : S _{ a } \cap T _{ n }=\phi\right\}\) is.
- A \(0\)
- B \(2\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(D) \(4\)
Step-by-step Solution
Detailed explanation
\(Bonus\) \(S_{n}:|z-(3-2 i)|=\frac{n}{4}\) 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 \(S_{ a } \cap T _{ n }=\phi\) Both circles do not intersect each other…
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