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. The equation of circle C is

Options

A. `(x-2sqrt3)^(2)+(x-1)^(2)=1`
B. `(x-2sqrt3)^(2)+(y-(1)/(2))^(2)=1`
C. `(x-sqrt3)^(2)+(y+1)^(2) = 1`
D. `(x-sqrt3)^(2)+(y-1)^(2) = 1`

Answer

D

Explanation

Let centre of circle C be (h,k).
Then, `|(sqrt3h+k-6)/(sqrt(3+1))|=1`
`rArr sqrt3h+k-6=2,-2`
`rArr sqrt3h+k=4" "...(i)`
[rejecting 2 because origin and centre of C are on the same side of PQ]
The point `(sqrt3,1)` satisfies Eq. (i).
`therefore` Equation of circle C is `(x-sqrt3)^(2)+(y-1)^(2)=1`.
Clearly, point E and F satisfy the equation given in option (d).