MCQ
If the points $(a,b),\,(a',b')$and $(a - a',b - b')$are collinear, then
- ✓$ab' = a'b$
- B$ab = a'b'$
- C$aa' = bb'$
- D${a^2} + {b^2} = 1$
$ \Rightarrow \,\,\,\frac{{a - 2a'}}{{a' - a}} = \frac{{b - 2b'}}{{b' - b}}$
$ \Rightarrow \,\,\frac{a}{{a'}} = \frac{b}{{b'}}\,\,$
$\Rightarrow \,\,ab' = a'b.$
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$M =\left\{( x , y ) \in R \times R : x ^2+ y ^2 \leq r ^2\right\},$
where $r >0$. Consider the geometric progression $a _{ n }=\frac{1}{2^{ n -1}}, n =1,2,3, \ldots$. Let $S _0=0$ and, for $n \geq 1$, let $S _{ n }$ denote the sum of the first $n$ terms of this progression. For $n \geq 1$, let $C_n$ denote the circle with center $\left(S_{n-1}, 0\right)$ and radius $a _{ n }$, and $D _{ n }$ denote the circle with center $\left( S _{ n -1}, S _{ n -1}\right)$ and radius $a _{ n }$.
($1$) 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
$(A)$ $k +2 l=22$ $(B)$ $2 k +l=26$ $(C)$ $2 k +3 l=34$ $(D)$ $3 k +2 l=40$
($2$) Consider $M$ with $r =\frac{\left(2^{199}-1\right) \sqrt{2}}{2^{158}}$. The number of all those circles $D _{ a }$ that are inside $M$ is
$(A) 198$ $(B) 199$ $(C) 200$ $(D) 201$
Give the answer or qution ($1$) and ($2$)