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 are

Options

A. `(x-2sqrt(3))^(2)+(y-1)^(2)=1`
B. `(x-2sqrt(3))^(2)+(y+(1)/(2))^(2)=1`
C. `(x-sqrt(3))^(2)+(y+1)^(2)=1`
D. `(x-sqrt(3))^(2)+(y-1)^(2)=1`

Answer

D

Explanation

Let A(h,k) be the centre of the circle. Then, `AD_|_PQ` and AD=1

`rArr (2k-3)/(2h-3sqrt(3))xx(-sqrt(3))=-1 and |(sqrt(3)h+k-6)/(sqrt(3)+1)|=1`
`rArr 2sqrt(3)k-3sqrt(3)=2h-3sqrt(3) and |sqrt(3)h+k-6|=2`
`rArr 2h-2sqrt( 3)k=0 and sqrt(3)h+k-6 = 2 or sqrt(3)h+k-6=-2`
`rArr h=sqrt(3)k and sqrt(3)h+k-6=-2`
[`:.`( Origin and (h, k) lie on the same side of `sqrt(3)x+y-6=0), (. :. sqrt(3)h+k-6 lt 0)]`
`rArr h=sqrt(3)k and sqrt(3)h+k=4`
`rArr h=sqrt(3), k=1`
Hence, the equation of circle C is `(x-sqrt(3))^(2)+(y-1)^(2)=1`