Number of elements where n belongs to N and Sn Intersection Tn is phi

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 $\left\{ {n \in N:{S_n} \cap {T_n} = \phi } \right\}$ is

(A) 0
(B) 2
(C) 3
(D) 4

Solution:

Tip for solving this question:

Firstly Identify the given sets

According to the set type, apply intersection properties to find no. of elements.

Step 1 of 2:

Let z = x + iy

Given, ${S_n} = \left\{ {z \in C:|z - 3 + 2i| = \frac{n}{4}} \right\}$

$\begin{gathered} |z - 3 + 2i| = \frac{n}{4} \hfill \\ |x + iy - 3 + 2i| = \frac{n}{4} \hfill \\ |(x - 3) + (y + 2)i| = \frac{n}{4} \hfill \\ {S_n}:{(x - 3)^2} + {(y + 2)^2} = {\left( {\frac{n}{4}} \right)^2} \hfill \\ \end{gathered}$

Represents a circle with centre ${C_1}(3, - 2) \text{ and radius } {r_1} = \frac{n}{4}$

Also, ${T_n} = \left\{ {z \in C:|z - 2 + 3i| = \frac{1}{n}} \right\}$

$\begin{gathered} |z - 2 + 3i| = \frac{1}{n} \hfill \\ |x + iy - 2 + 3i| = \frac{1}{n} \hfill \\ |(x - 2) + (y + 3)i| = \frac{1}{n} \hfill \\ {T_n}:{(x - 2)^2} + {(y + 3)^2} = {\left( {\frac{1}{n}} \right)^2} \hfill \\ \end{gathered}$

Represents a circle with centre ${C_2}(2, - 3) \text{ and radius } {r_2} = \frac{1}{n}$

Step 2 of 2:

We have $\left\{ {n \in N:{S_n} \cap {T_n} = \phi } \right\}$

$\text{ i.e.} {S_n} \cap {T_n} = \phi$

So, distance between centre ${C_1}(3, - 2) \text{ and } {C_2}(2, - 3)$ > ${r_1} + {r_2}$

$\Rightarrow \sqrt {{{(2 - 3)}^2} + {{( - 1 - (2))}^2}} > \frac{n}{4} + \frac{1}{n}$

{ $\because \text{ Distance between } ({x_1},{x_2})\& ({y_1},{y_2}) = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}$ }

$\begin{gathered} \Rightarrow \sqrt {1 + 1} > \frac{{{n^2} + 4}}{{4n}} \hfill \\ \Rightarrow \sqrt 2 > \frac{{{n^2} + 4}}{{4n}} \hfill \\ \Rightarrow 4n\sqrt 2 > {n^2} + 4 \hfill \\ \Rightarrow {n^2} - 4n\sqrt 2 + 4 < 0 \hfill \\ \Rightarrow {n^2} - 4n\sqrt 2 + 8 - 4 < 0 \hfill \\ \Rightarrow {(n - 2\sqrt 2 )^2} - {(2)^2} < 0 \hfill \\ \Rightarrow (n - 2\sqrt 2 + 2)(n - 2\sqrt 2 - 2) < 0 \hfill \\ \Rightarrow (n - 2\sqrt 2 - 2) < 0 \hfill \\ \Rightarrow n < 2 + 2\sqrt 2 \hfill \\ \end{gathered}$

Here, n = 1, 2, 3, 4 only satisfies the above equation.

Therefore, No. of elements in sets = 4

Final Answer:

Hence, Option (D) is correct.

Solution:

Step 1 of 2:

Step 2 of 2:

Final Answer:

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