Let M = \ (x,y) \in R \times R:x^2 + y^2 \le r^2\ , where r > 0. Consider the geometric progression a_n = 1 \over 2^n - 1 , n = 1, 2, 3, ...... . Let S0 = 0 and for n \ge 1, let Sn denote the sum of the first n terms of this progression. For n \ge 1, let Cn denote the circle with center (Sn - 1, 0) and radius an, and Dn denote the circle with center (Sn - 1, Sn - 1) and radius an. Consider M with r = (2^199 - 1)\sqrt 2 \over 2^198 . The number of all those circles Dn that are inside M is

Circle PYQ — JEE Advanced 2021 · Shift 2 · Q33 with Solution

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Correct answer: NONE OF THE OPTIONS IS CORRECT; THE NUMBER OF CIRCLES $D_N$ INSIDE $M$ IS INFINITE.

Given data

The GP is $a_n=\frac{1}{2^{n-1}},\qquad n\ge 1.$

Its first $n$ terms sum is

Since this is a GP with first term $1$ and ratio $\frac12$ ,

Hence $S_n-1=1-\frac{1}{2^{n-1}}=1-a_n.$
Circle $D_n$

Circle $D_n$ has center $\bigl(S_n-1,\; S_n-1\bigr)=\bigl(1-a_n,\;1-a_n\bigr)$ and radius $a_n.$
Condition for $D_n$ to lie inside $M$

The set $M=\{(x,y):x^2+y^2\le r^2\}$ is the disk centered at origin with radius $r$ .

For the whole circle $D_n$ to be inside $M$ , we need $\text{distance of center from origin} + \text{radius} \le r.$

Distance of center of $D_n$ from origin: $\sqrt{(1-a_n)^2+(1-a_n)^2}=\sqrt2\,(1-a_n).$

Therefore the required condition is $\sqrt2(1-a_n)+a_n\le r.$
Substitute $a_n=2^{-(n-1)}$ and given $r$

We are given $r=\frac{(2^{199}-1)\sqrt2}{2^{198}}=\sqrt2\left(2-\frac{1}{2^{198}}\right).$

Let $t=a_n=\frac{1}{2^{n-1}}.$ Then the inclusion condition becomes $\sqrt2(1-t)+t\le \sqrt2\left(2-\frac1{2^{198}}\right).$

Rearranging, $\sqrt2 -\sqrt2 t + t \le 2\sqrt2-\frac{\sqrt2}{2^{198}}$ $t(1-\sqrt2)\le \sqrt2-\frac{\sqrt2}{2^{198}}.$

Since $1-\sqrt2<0$ , it is cleaner to compare directly by testing the boundary value suggested by the power of $2$ .
Check $n=199$

For $n=199$ , $a_{199}=\frac{1}{2^{198}}.$ Then
$=\sqrt2\left(1-\frac1{2^{198}}\right)+\frac1{2^{198}}.$
Compare with

Their difference is
$=\left(2\sqrt2-\frac{\sqrt2}{2^{198}}\right)-\left(\sqrt2-\frac{\sqrt2}{2^{198}}+\frac1{2^{198}}\right) =\sqrt2-\frac1{2^{198}}>0.$
So $D_{199}$ is inside $M$ .
Check $n=200$

For $n=200$ , $a_{200}=\frac1{2^{199}}.$ Then

Now
$=\left(2\sqrt2-\frac{\sqrt2}{2^{198}}\right)-\left(\sqrt2-\frac{\sqrt2}{2^{199}}+\frac1{2^{199}}\right) =\sqrt2-\frac{\sqrt2}{2^{199}}-\frac{\sqrt2}{2^{198}}-\frac1{2^{199}}>0.$
So $D_{200}$ is also inside $M$ .

In fact, as $n$ increases, $a_n$ decreases and the center approaches $(1,1)$ , so $\sqrt2(1-a_n)+a_n$ decreases because its coefficient of $a_n$ is $1-\sqrt2<0$ . Hence if $D_{200}$ is inside, then all $D_n$ for $n\ge 200$ are also inside.
Conclusion

Therefore, the number of circles $D_n$ inside $M$ is not finite; certainly it is more than $201$ .

So none of the given options is correct.
Comparison with stored answer

The stored answer is B: 199, but our derivation shows that infinitely many circles $D_n$ are inside $M$ . Hence the stored answer is incorrect.

It is possible that the intended question was about circles $C_n$ instead of $D_n$ , or some additional restriction on $n$ was omitted.

More from Circle

View written solutionFree

Correct answer: NONE OF THE OPTIONS IS CORRECT; THE NUMBER OF CIRCLES $D_N$ INSIDE $M$ IS INFINITE.

Given data

The GP is $a_n=\frac{1}{2^{n-1}},\qquad n\ge 1.$

Its first $n$ terms sum is

Since this is a GP with first term $1$ and ratio $\frac12$ ,

Hence $S_n-1=1-\frac{1}{2^{n-1}}=1-a_n.$
Circle $D_n$

Circle $D_n$ has center $\bigl(S_n-1,\; S_n-1\bigr)=\bigl(1-a_n,\;1-a_n\bigr)$ and radius $a_n.$
Condition for $D_n$ to lie inside $M$

The set $M=\{(x,y):x^2+y^2\le r^2\}$ is the disk centered at origin with radius $r$ .

For the whole circle $D_n$ to be inside $M$ , we need $\text{distance of center from origin} + \text{radius} \le r.$

Distance of center of $D_n$ from origin: $\sqrt{(1-a_n)^2+(1-a_n)^2}=\sqrt2\,(1-a_n).$

Therefore the required condition is $\sqrt2(1-a_n)+a_n\le r.$
Substitute $a_n=2^{-(n-1)}$ and given $r$

We are given $r=\frac{(2^{199}-1)\sqrt2}{2^{198}}=\sqrt2\left(2-\frac{1}{2^{198}}\right).$

Let $t=a_n=\frac{1}{2^{n-1}}.$ Then the inclusion condition becomes $\sqrt2(1-t)+t\le \sqrt2\left(2-\frac1{2^{198}}\right).$

Rearranging, $\sqrt2 -\sqrt2 t + t \le 2\sqrt2-\frac{\sqrt2}{2^{198}}$ $t(1-\sqrt2)\le \sqrt2-\frac{\sqrt2}{2^{198}}.$

Since $1-\sqrt2<0$ , it is cleaner to compare directly by testing the boundary value suggested by the power of $2$ .
Check $n=199$

For $n=199$ , $a_{199}=\frac{1}{2^{198}}.$ Then
$=\sqrt2\left(1-\frac1{2^{198}}\right)+\frac1{2^{198}}.$
Compare with

Their difference is
$=\left(2\sqrt2-\frac{\sqrt2}{2^{198}}\right)-\left(\sqrt2-\frac{\sqrt2}{2^{198}}+\frac1{2^{198}}\right) =\sqrt2-\frac1{2^{198}}>0.$
So $D_{199}$ is inside $M$ .
Check $n=200$

For $n=200$ , $a_{200}=\frac1{2^{199}}.$ Then

Now
$=\left(2\sqrt2-\frac{\sqrt2}{2^{198}}\right)-\left(\sqrt2-\frac{\sqrt2}{2^{199}}+\frac1{2^{199}}\right) =\sqrt2-\frac{\sqrt2}{2^{199}}-\frac{\sqrt2}{2^{198}}-\frac1{2^{199}}>0.$
So $D_{200}$ is also inside $M$ .

In fact, as $n$ increases, $a_n$ decreases and the center approaches $(1,1)$ , so $\sqrt2(1-a_n)+a_n$ decreases because its coefficient of $a_n$ is $1-\sqrt2<0$ . Hence if $D_{200}$ is inside, then all $D_n$ for $n\ge 200$ are also inside.
Conclusion

Therefore, the number of circles $D_n$ inside $M$ is not finite; certainly it is more than $201$ .

So none of the given options is correct.
Comparison with stored answer

The stored answer is B: 199, but our derivation shows that infinitely many circles $D_n$ are inside $M$ . Hence the stored answer is incorrect.

It is possible that the intended question was about circles $C_n$ instead of $D_n$ , or some additional restriction on $n$ was omitted.

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