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JEE Advance - Mathematics (2008 - Paper 1 Offline - No. 15)

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.
The equation of circle C is
$${\left( {x\, - 2\sqrt 3 \,} \right)^2} + {(y - 1)^2} = 1$$
$${\left( {x\, - 2\sqrt 3 \,} \right)^2} + {(y + {1 \over 2})^2} = 1$$
$${\left( {x\, - \sqrt 3 \,} \right)^2} + {(y + 1)^2} = 1$$
$${\left( {x\, - \sqrt 3 \,} \right)^2} + {(y - 1)^2} = 1$$

Explanation

Let centre of circle C be (h, k) then,

$$\left| {{{\sqrt 3 h + k - 6} \over {\sqrt {3 + 1} }}} \right| = 1$$

$$\sqrt 3 h + k - 6 = 2, - 2$$

$$\sqrt 3 h + k = 4$$ (rejecting 2 because origin and centre of point $$\left( {\sqrt 3 ,1} \right)$$, satisfies eq.)

eq. of circle (C) is

$${\left( {x - \sqrt 3 } \right)^2} + {(y - 1)^2} = 1$$

Clearly point E and F satisfy the equation in given option D.

As $${\left( {x - 2\sqrt 3 } \right)^2} + {(y - 1)^2} = 1$$ not possible.

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