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.Points E and F are given by

Options

A. a)`((sqrt(3))/(2),(3)/(2)),(sqrt(3),0)`
B. b)`((sqrt(3))/(2),(1)/(2)),(sqrt(3),0)`
C. c)`((sqrt(3))/(2),(3)/(2)),((sqrt(3))/(2),(1)/(2))`
D. d)`((3)/(2),(sqrt(3))/(2)),((sqrt(3))/(2),(1)/(2))`

Answer

1

Explanation

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`