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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 `sqrt3 x+ y -6 = 0` and the point D is (3 sqrt3/2, 3/2). Further, it is given that the origin and the centre of C are on the same side of the line PQ. (1)The equation of circle C is (2)Points E and F are given by (3)Equation of the sides QR, RP areA. `((sqrt(3))/(2),(3)/(2)),(sqrt(3),0)`B. `((sqrt(3))/(2),(1)/(2)),(sqrt(3),0)`C. `((sqrt(3))/(2),(3)/(2)),((sqrt(3))/(2),(1)/(2))`D. `((3)/(2),(sqrt(3))/(2)),((sqrt(3))/(2),(1)/(2))` |
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Answer» Correct Answer - 1 By simple geometry `PD= sqrt(3) ( Delta PQR` is equilateral ) Equation of PQ in parametric form is `(x-(3sqrt(3))/(2))/( cos 120^(@))=(y-(3)/(2))/(sin 120^(@))=r` or `(x-(3sqrt(3))/(2))/( -(1)/(2))=(y-(3)/(2))/((sqrt(3))/(2))=r` Now points at distance `sqrt(3)` from point D on line PQ are given by `((3sqrt(3))/(2) +- sqrt(3) (-(1)/(2)),(3)/(2)+- sqrt(3) ((sqrt(3))/(2)))` or `(2 sqrt(3),0)` and `( sqrt(3),3)` Point C divides the join of P and E in the ratio `2:1` . Similarly, C divides join of Q and F in the ratio `2:1` |
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