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JEE Advance - Mathematics (2008 - Paper 1 Offline - No. 17)

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 $$\sqrt 3 x\, + \,y\, - \,6 = 0$$ and the point D is $$\left( {{{3\,\sqrt 3 } \over 2},\,{3 \over 2}} \right)$$. Further, it is given that the origin and the centre of C are on the same side of the line PQ.
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 $$\sqrt 3 x\, + \,y\, - \,6 = 0$$ and the point D is $$\left( {{{3\,\sqrt 3 } \over 2},\,{3 \over 2}} \right)$$. Further, it is given that the origin and the centre of C are on the same side of the line PQ.
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 $$\sqrt 3 x\, + \,y\, - \,6 = 0$$ and the point D is $$\left( {{{3\,\sqrt 3 } \over 2},\,{3 \over 2}} \right)$$. Further, it is given that the origin and the centre of C are on the same side of the line PQ.
Equations of the sides QR, RP are
$$y = {2 \over {\sqrt 3 }}\,x + \,1,\,\,y = \, - {2 \over {\sqrt 3 }}\,x - 1$$
$$y = {1 \over {\sqrt 3 }}\,x,\,\,y = \,0$$
$$y = {{\sqrt 3 } \over 2}\,x + \,1,\,\,y = \, - {{\sqrt 3 } \over 2}\,x - 1$$
$$y = \sqrt 3 \,x,\,\,y = \,0$$

Explanation

Equation of PR is X-axis, that is $$y = 0$$ and the equation of side QR is

$$\left( {y - {3 \over 2}} \right) = \sqrt 3 \left( {x - {{\sqrt 3 } \over 2}} \right)$$

$$y = \sqrt 3 x$$

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