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. a)`(x-2sqrt(3))^(2)+(y-1)^(2)=1`
B. b)`(x-2sqrt(3))^(2)+(y+(1)/(2))^(2)=1`
C. c)`(x-sqrt(3))^(2)+(y+1)^(2)=1`
D. d)`(x-sqrt(3))^(2)+(y-1)^(2)=1`

Answer

4

Explanation

Equation of `PQ` is `L_(1) : sqrt(3) x +y-6=0` (1)
Slope of PQ is `- sqrt(3)`
`:. `Slope of CD is `(1)/(sqrt(3))`.
`:. `Equation of line CD in parametric from is
`(x-(3sqrt(3))/(2))/(cos 30^(@))=(y-(3)/(2))/(sin 30^(@))=r`
or `(x-(3sqrt(3))/(2))/((sqrt(3))/(2))=(y-(3)/(2))/((1)/(2))=r`
For centre C, `r = +- 1`
Two possible coordinates of centre are `(2sqrt(3),2),(sqrt(3),1)`
According to the question, center lies on the same side where origin lies with respect to line PQ.
Now `L_(1) (0,0) = -6 lt0`
And `L_(1) ( sqrt(3),1) = 3+1-6 lt0`
So Centre C must be `( sqrt(3), 1)`
Hence, equation of the circle is `(x-sqrt(3))^(2)+(y-1)^(2) =1`