Question

Let $M=\left\{(x,y)\in R\times R:x^2+y^2\le r^2\right\}$, where $r>0$. Consider the geometric progression $a_n=\frac{1}{2^{n-1}},n=1,2,3,\dots .$ Let $S_0=0$ and, for $n\ge 1$, let $S_n$ denote the sum of the first $n$ terms of this progression. For $n\ge 1$ let $C_n$ denote the circle with center $(S_{n-1},0)$ and radius $a_n$ and $D_n$ denote the circle with center $(S_{n-1},S_{n-1})$ and radius $a_n$.

Consider $M$ with $r=\frac{1025}{513}$. 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

• A. $k+2l=22$
• B. $2k+l=26$
• C. $2k+3l=34$
• D. $3k+2l=40$

Answer 1

The correct option is D $3k+2l=40$

$\because a_n=\frac{1}{2^{n-1}}$ and $S_n=2(1-\frac{1}{2^n})$
For circles $C_n$ to be inside $M$.
$S_{n-1}+a_n<\frac{1025}{513}$
$\Rightarrow S_n<\frac{1025}{513}$
$\Rightarrow 1-\frac{1}{2^n}<\frac{1025}{1026}=1-\frac{1}{1026}$
$\Rightarrow 2^n<1026$
$\Rightarrow n\le 10$
$\therefore$ Number of circles inside $M$ be $10=k$
Clearly alternate circles do not intersect each other i.e., $C_1,C_3,C_5,C_7,C_9$ do not intersect each other as well as $C_2,C_4,C_6,C_8$ and $C_{10}$ do not intersect each other hence maximum $5$ set of circles do not intersect each other.
$\therefore l=5$
$\therefore 3k+2l=40$