Test Index

IIT JEE Advanced 2008 Paper 1

Paragraph for Questions 64–66:
A circle C of radius 1 is inscribed in an equilateral triangle PQR. The points of contact of C with the sides PQ, QR, RP are D, E, F, respectively. The line PQ is given by the equation 3x + y - 6 = 0 and the point D is (

3√3

). Further, it is given that the origin and the centre C are on the same side of the line PQ.

Question : 64

Total: 69

The equation of the circle is

(x−2√3)2+(y−1)2 = 1
(x−2√3)2+(y+
1
2
)2 = 1
(x−√3)2+(y+1)2 = 1
(x−√3)2+(y−1)2 = 1

Validate

Go to Question:

Prev Question

Next Question