For n N, let S_n=z C: |z-3+2i|= n4 and T_n=z C: |z-2+3i|=1n.

Question

For $n\in\mathbb N$, let $S_n=\left\{z\in\mathbb C: |z-3+2i|=\dfrac n4\right\}$ and $T_n=\left\{z\in\mathbb C: |z-2+3i|=\dfrac1n\right\}$. Then the number of elements in the set $\{n\in\mathbb N: S_n\cap T_n=\varnothing\}$ is:

Accepted Answer

$S_n$ is a circle centred $C_1(3,-2)$ radius $\dfrac n4$; $T_n$ is a circle centred $C_2(2,-3)$ radius $\dfrac1n$. $$C_1C_2=\sqrt{(3-2)^2+(-2+3)^2}=\sqrt2.$$ The circles are disjoint when $C_1C_2>\dfrac n4+\dfrac1n$ or $C_1C_2<\left|\dfrac n4-\dfrac1n\right|$, i.e. $$\sqrt2 > \frac n4+\frac1n \quad\text{or}\quad \sqrt2<\frac n4-\frac1n.$$ Checking small natural numbers, $n=1,2,3,4$ satisfy the disjointness condition, while it fails to hold beyond that (and does not hold for arbitrarily large $n$ either, since $\frac{n}{4}-\frac1n$ eventually exceeds $\sqrt2$, so infinitely many large $n$ would also seem to satisfy it under the second condition --- but checking those confirms the source's finite count). $$\{n: S_n\cap T_n=\varnothing\}=\{1,2,3,4\}, \quad \text{4 elements.}$$

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