JEE Advanced 2021 Paper 2 Q13 Mathematics Coordinate Geometry Circles Hard

JEE Advanced 2021 Paper 2 · Q13 · Circles

Consider $M$ with $r = \dfrac{1025}{513}$. Let $k$ be the number of all those circles $C_n$ that are inside $M$. Let $l$ be the maximum possible number of circles among these $k$ circles such that no two circles intersect. Then

  1. A. $k + 2l = 22$
  2. B. $2k + l = 26$
  3. C. $2k + 3l = 34$
  4. D. $3k + 2l = 40$
Reveal answer + step-by-step solution

Correct answer:D

Solution

$a_n = \dfrac{1}{2^{n-1}}$, $S_{n-1} = 2(1 - 1/2^{n-1})$. Center of $C_n$ is $(S_{n-1}, 0)$, radius $a_n$. $C_n$ lies inside $M$ iff $S_{n-1} + a_n \le r$, i.e. $2 - \dfrac{1}{2^{n-1}} \le \dfrac{1025}{513}$, giving $\dfrac{1}{2^{n-1}} \ge \dfrac{1}{513}$, so $2^{n-1} \le 513$, hence $n \le 10$. So $k = 10$. Non-intersecting subfamily: pick every other circle (e.g. $C_1, C_3, C_5, C_7, C_9$ or $C_2, C_4, C_6, C_8, C_{10}$), giving $l = 5$. Hence $3k + 2l = 30 + 10 = 40$. Option (D).

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