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JEE Advanced 2008Paper 1
Exam
JEE Advanced
Year
2008
Paper
Paper 1
Subject
Mathematics
Chapter
Circle

JEE Advanced 2008 · Paper 1 · Mathematics · Circle

Paragraph: A circle C of radius 1 is inscribed in an equilateral P...

Paragraph: A circle CC of radius 1 is inscribed in an equilateral △PQR\triangle P Q R. The points of contact of CC with the sides PQ,QR,RPP Q, Q R, R P are D,E,FD, E, F respectively. The line PQP Q is given by the equation 3x+y−6=0\sqrt{3} x+y-6=0 and the point DD is \left(\frac{3 \sqrt{3}}{2}, \frac{3}{2}\right).Further,itisgiventhattheoriginandthecentreof. Further, it is given that the origin and the centre ofCareonthesamesideofthelineare on the same side of the lineP Q.<strong>Question:</strong>Theequationofcircle.<strong>Question:</strong> The equation of circleC is

Options

  1. A(x−23)2+(y−1)2=1(x-2 \sqrt{3})^2+(y-1)^2=1
  2. B(x−23)2+(y+12)2=1(x-2 \sqrt{3})^2+\left(y+\frac{1}{2}\right)^2=1
  3. C(x−3)2+(y+1)2=1(x-\sqrt{3})^2+(y+1)^2=1
  4. D(x−3)2+(y−1)2=1(x-\sqrt{3})^2+(y-1)^2=1

Correct answer

D. (x−3)2+(y−1)2=1(x-\sqrt{3})^2+(y-1)^2=1

Step-by-step solution

Let centre of circle CC be (h,k)(h, k). Then, ∣3h+k−63+1∣=1\left|\frac{\sqrt{3} h+k-6}{\sqrt{3+1}}\right|=1 \Rightarrow \quad 3\sqrt{3}h+k-6=2,-2 \Rightarrow \quad 3\sqrt{3}h+k=4 (Rejecting 2 because origin and centre of CC are on the same side of PQP Q ) The point (3,1)(\sqrt{3}, 1) satisfies Eq. (i). ∴\therefore Equation of circle CC is (x−3)2+(y−1)2=1(x-\sqrt{3})^2+(y-1)^2=1.
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