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JEE Advanced 2008Paper 1
Exam
JEE Advanced
Year
2008
Paper
Paper 1
Subject
Mathematics
Chapter
Straight Lines

JEE Advanced 2008 · Paper 1 · Mathematics · Straight Lines

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>Points.<strong>Question:</strong> PointsE and F are given byParagraph: A circle C of radius 1 is inscribed in an equilateral P Q R . The points of contact of C with the sides P Q,

Options

  1. A(32,32),(3,0)\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),(\sqrt{3}, 0)
  2. B(32,12),(3,0)\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right),(\sqrt{3}, 0)
  3. C(32,32),(32,12)\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)
  4. D(32,32),(32,12)\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right),\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)

Correct answer

A. (32,32),(3,0)\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right),(\sqrt{3}, 0)

Step-by-step solution

Slope of line joining centre of circle to point DD =32−1332−3=13=\frac{\frac{3}{2}-1}{\frac{3 \sqrt{3}}{2}-\sqrt{3}}=\frac{1}{\sqrt{3}} [makes an angle 30∘30^{\circ} with XX-axis] ∴\therefore Points EE and FF will make angle 150∘150^{\circ} and −90∘-90^{\circ} with XX-axis. Paragraph: A circle C of radius 1 is inscribed in an equilateral P Q R . The points of contact of C with the sides P Q,
∴E\therefore E and FF are given by and ∴\therefore \begin{aligned} \frac{x-\sqrt{3}}{\cos 150^{\circ}} & =\frac{y-1}{\sin 150^{\circ}}=1 \\ \frac{x-\sqrt{3}}{\cos \left(-90^{\circ}\right)} & =\frac{y-1}{\sin \left(-90^{\circ}\right)}=1 \\ E & =\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right) \\ and F & =(\sqrt{3}, 0) \end{aligned}
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