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JEE Advanced 2006
Exam
JEE Advanced
Year
2006
Subject
Mathematics
Chapter
Circle

JEE Advanced 2006 · Mathematics · Circle

Paragraph: Read the following passage and answer the questions. Let...

Paragraph: Read the following passage and answer the questions. Let ABCDA B C D be a square of side length 2 units. C2C_2 is the circle through vertices A,B,C,DA, B, C, D and C1C_1 is the circle touching all the sides of square ABCDA B C D. LL is the line through AA.Question: If PP is a point on C1C_1 and QQ is a point on C2C_2, then PA2+PB2+PC2+PD2QA2+QB2+QC2+QD2\frac{P A^2+P B^2+P C^2+P D^2}{Q A^2+Q B^2+Q C^2+Q D^2} is equal toParagraph: Read the following passage and answer the questions. Let A B C D be a square of side length 2 units. C_2 isParagraph: Read the following passage and answer the questions. Let A B C D be a square of side length 2 units. C_2 is

Options

  1. A0.750.75
  2. B1.251.25
  3. C1
  4. D0.50.5

Correct answer

A. 0.750.75

Step-by-step solution

\text{ Here, equation of }C2:(x−1)2+(y−1)2=(C_2:(x-1)^2+(y-1)^2=(\sqrt{2})^22\text{ and }C1:(x−1)2+(y−1)2=(1)2C_1:(x-1)^2+(y-1)^2=(1)^2∴P(1+cos⁡θ,1+sin⁡θ) and Q(1+2sin⁡θ,1+2sin⁡θ)∴PA2+PB2+PC2+PD2={(1+cos⁡θ)2+(1+sin⁡θ)2}+{(cos⁡θ−1)2+(1+sin⁡θ)2}+{(1+cos⁡θ)2+(1−sin⁡θ)2}+{(1−cos⁡θ)2+(1−sin⁡θ)2}=16∴ΣQA2ΣPA2=1216=0.75\begin{aligned}& \therefore P(1+\cos \theta,1+\sin \theta)\\ & \text{ and }Q(1+\sqrt{2} \sin \theta,1+\sqrt{2} \sin \theta)\\ & \therefore P A²+P B²+P C²+P D² \\ & =\left\{(1+\cos \theta)^2+(1+\sin \theta)^2\right\} \\ & +\left\{(\cos \theta-1)^2+(1+\sin \theta)^2\right\} \\ & +\left\{(1+\cos \theta)^2+(1-\sin \theta)^2\right\} \\ & +\left\{(1-\cos \theta)^2+(1-\sin \theta)^2\right\} \\ & =16 \\ & \therefore \quad \frac{\Sigma Q A^2}{\Sigma P A^2}=\frac{12}{16}=0.75 \\ & \end{aligned}$Hence (a) is the correct answer. Paragraph: Read the following passage and answer the questions. Let A B C D be a square of side length 2 units. C_2 is
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