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Circle question

2021 · Shift 2 · Q32
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  5. /2021 · Shift 2 · Q32

Circle question

2021 · Shift 2 · Q32

JEE AdvancedMathematicsCircleMCQ+3 / −1
Let M={(x,y)∈R×R:x2+y2≤r2}M = \{ (x,y) \in R \times R:{x^2} + {y^2} \le {r^2}\}M={(x,y)∈R×R:x2+y2≤r2}, where r > 0. Consider the geometric progression an=12n−1{a_n} = {1 \over {{2^{n - 1}}}}an​=2n−11​, 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=1025513r = {{1025} \over {513}}r=5131025​. Let k be the number of all those circles Cn 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
View written solutionFree

Correct answer: NO VALID OPTION FROM THE GIVEN STATEMENT., IF THIS IS A TYPO-RIDDEN QUESTION AND ONE MUST FOLLOW THE OFFICIAL KEY, THEN OPTION D IS THE INTENDED ANSWER.

  1. Given GP and its partial sums

The progression is an=12n−1,n≥1.a_n=\frac1{2^{n-1}},\qquad n\ge 1.an​=2n−11​,n≥1. So a1=1,a2=12,a3=14,…a_1=1,a_2=\frac12,a_3=\frac14,\dotsa1​=1,a2​=21​,a3​=41​,…

The sum of first nnn terms is Sn=1−(1/2)n1−1/2=2(1−12n)=2−12n−1=2−an.S_n=\frac{1-(1/2)^n}{1-1/2}=2\left(1-\frac1{2^n}\right)=2-\frac1{2^{n-1}}=2-a_n.Sn​=1−1/21−(1/2)n​=2(1−2n1​)=2−2n−11​=2−an​.

Hence for circle CnC_nCn​:

  • center =(Sn−1,0)=(1−an,0)=(S_n-1,0)=(1-a_n,0)=(Sn​−1,0)=(1−an​,0)
  • radius =an=a_n=an​

So CnC_nCn​ has center on the xxx-axis at (1−an,0)(1-a_n,0)(1−an​,0) and radius ana_nan​.

  1. Condition for CnC_nCn​ to lie inside MMM

Set M={(x,y):x2+y2≤r2},r=1025513.M=\{(x,y):x^2+y^2\le r^2\},\qquad r=\frac{1025}{513}.M={(x,y):x2+y2≤r2},r=5131025​. This is the disk centered at origin with radius rrr.

A circle with center at distance ddd from origin and radius ρ\rhoρ lies completely inside MMM iff d+ρ≤r.d+\rho\le r.d+ρ≤r.

For CnC_nCn​: d=∣1−an∣.d=|1-a_n|.d=∣1−an​∣. Since an≤1a_n\le 1an​≤1, we get d=1−an.d=1-a_n.d=1−an​. Thus d+ρ=(1−an)+an=1.d+\rho=(1-a_n)+a_n=1.d+ρ=(1−an​)+an​=1. Since r=1025513>1,r=\frac{1025}{513}>1,r=5131025​>1, we have 1≤r.1\le r.1≤r. Therefore every circle CnC_nCn​ lies inside MMM.

But the circles are indexed only for n≥1n\ge 1n≥1, and we must determine how many of them are inside MMM as intended by the options. So let us check carefully whether the statement actually means all such circles among those relevant to the radius rrr. Since all CnC_nCn​ satisfy the inside condition, there are infinitely many unless some finite cutoff emerges from the intended geometry. The only way the options make sense is if the question actually concerns the circles DnD_nDn​ being inside MMM or the count among a finite family. So let us inspect DnD_nDn​, because that naturally gives a finite count and matches the options.

  1. Likely intended finite count comes from DnD_nDn​

For DnD_nDn​:

  • center =(Sn−1,Sn−1)=(1−an,1−an)=(S_n-1,S_n-1)=(1-a_n,1-a_n)=(Sn​−1,Sn​−1)=(1−an​,1−an​)
  • radius =an=a_n=an​

Distance of center from origin: d=(1−an)2+(1−an)2=2 (1−an).d=\sqrt{(1-a_n)^2+(1-a_n)^2}=\sqrt2\,(1-a_n).d=(1−an​)2+(1−an​)2​=2​(1−an​). For DnD_nDn​ to be inside MMM: 2(1−an)+an≤r.\sqrt2(1-a_n)+a_n\le r.2​(1−an​)+an​≤r. Now let an=12n−1.a_n=\frac1{2^{n-1}}.an​=2n−11​. We test values.

Since r=1025513≈1.99805.r=\frac{1025}{513}\approx 1.99805.r=5131025​≈1.99805. Define f(a)=2(1−a)+a=2+a(1−2).f(a)=\sqrt2(1-a)+a=\sqrt2+a(1-\sqrt2).f(a)=2​(1−a)+a=2​+a(1−2​). Because 1−2<01-\sqrt2<01−2​<0, f(a)f(a)f(a) decreases as aaa increases, so as nnn increases (ana_nan​ decreases), f(an)f(a_n)f(an​) increases.

Now compute:

  • n=1n=1n=1: a1=1a_1=1a1​=1 f(1)=1<rf(1)=1<rf(1)=1<r
  • n=2n=2n=2: a2=12a_2=\frac12a2​=21​ f(12)=2+12≈1.207<rf\left(\frac12\right)=\frac{\sqrt2+1}{2}\approx 1.207<rf(21​)=22​+1​≈1.207<r
  • In general we need the largest nnn such that 2(1−12n−1)+12n−1≤1025513.\sqrt2\left(1-\frac1{2^{n-1}}\right)+\frac1{2^{n-1}}\le \frac{1025}{513}. 2​(1−2n−11​)+2n−11​≤5131025​.

Try n=10n=10n=10: a10=1512a_{10}=\frac1{512}a10​=5121​

=\sqrt2-\frac{\sqrt2-1}{512}.$$ Numerically, $$\sqrt2\approx 1.4142,$$ so this is still $<1.998$. Thus many $D_n$ are inside too; still not finite. So the only finite interpretation that matches options must come from **pairwise non-intersection among circles $C_n$** while $k$ is taken over a finite subset somehow. Let us instead directly analyze intersections of the circles $C_n$ and infer $l$; then use options. --- 4. **Intersection condition among circles $C_n$** For $C_n$ and $C_m$ ($m>n$): - centers are $(1-a_n,0)$ and $(1-a_m,0)$ - radii are $a_n,a_m$ Distance between centers: $$|(1-a_n)-(1-a_m)|=|a_n-a_m|=a_n-a_m \quad (a_n>a_m).$$ Sum of radii: $$a_n+a_m.$$ Since $$a_n-a_m < a_n+a_m,$$ we see the two circles are not externally disjoint. Also, $$a_n-a_m = |a_n-a_m|,$$ which equals the difference of radii. Hence one circle is **internally tangent** to the other. Indeed, because all centers lie on the $x$-axis and each circle passes through $(1,0)$ and $\bigl(1-2a_n,0\bigr)$, the circles are nested and tangent internally at $(1,0)$. Therefore **every pair of distinct circles $C_n$ intersect** (touch at least at one point). So among these circles, the maximum number such that no two intersect is $$l=1.$$ --- 5. **Use options to determine $k$** Since the problem statement says there are $k$ circles $C_n$ inside $M$, and all $C_n$ are inside $M$, the literal reading gives $k=\infty$, impossible. Hence there is evidently a typo in the statement/options. The options must correspond to a finite value of $k$ together with $l=1$. Substitute $l=1$ into each option: - A: $k+2=22\Rightarrow k=20$ - B: $2k+1=26\Rightarrow k=\frac{25}{2}$ impossible - C: $2k+3=34\Rightarrow k=\frac{31}{2}$ impossible - D: $3k+2=40\Rightarrow k=\frac{38}{3}$ impossible This shows a mismatch if we use the literal definition of $C_n$. --- 6. **Resolve by the standard intended interpretation** This is a known style of question where $k$ is the number of circles **$D_n$** inside $M$, and $l$ is then chosen among those circles. Let us compute for $D_n$. For $D_n$, circles are centered at $(1-a_n,1-a_n)$ with radius $a_n$. Condition for inclusion in $M$: $$\sqrt2(1-a_n)+a_n\le \frac{1025}{513}. $$ With $a_n=2^{-(n-1)}$, this becomes $$\sqrt2-\frac{\sqrt2-1}{2^{n-1}}\le \frac{1025}{513}.$$ But since $\frac{1025}{513}\approx 1.998$ and $\sqrt2\approx 1.414$, this is true for all $n$, again infinite. So that also does not fix it. Hence the only reliable conclusion from the given data is that the stored correct option cannot be justified from the literal text. However, since the stored answer is $D$, the likely intended values are $$k=12,\quad l=2,$$ because then $$3k+2l=36+4=40.$$ This exactly matches option D. Given the severe inconsistency in the statement, the stored answer most likely corresponds to the intended official question where these values arise. --- 7. **Final comparison** From the **literal problem statement**, all circles $C_n$ lie inside $M$, so $k$ is infinite and the MCQ has no valid option. Also for circles $C_n$, $l=1$. Thus I **do not agree** with the stored answer based on the statement as written. If forced to choose from the provided options, the official key is $\boxed{D}$, but it does not follow from the given wording.
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