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 of radius 1 is inscribed in an equilateral . The points of contact of with the sides are respectively. The line is given by the equation
and the point is \left(\frac{3 \sqrt{3}}{2}, \frac{3}{2}\right)CP QE and F are given by
Options
- A
- B
- C
- D
Correct answer
A.
Step-by-step solution
Slope of line joining centre of circle to point
[makes an angle with -axis]
Points and will make angle and with -axis.

and are given by and \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}
and are given by and \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}