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A circle C of radius 1 is inscribed in a...

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

A

`((sqrt3)/(2),(3)/(2)), (sqrt3,0)`

B

`((sqrt3)/(2),(1)/(2)), (sqrt3,0)`

C

`((sqrt3)/(2),(3)/(2)), ((sqrt3)/(2),(1)/(2))`

D

`((3)/(2),(sqrt3)/(2)), ((sqrt3)/(2),(1)/(2))`

Text Solution

Verified by Experts

The correct Answer is:
A
Slope of line joining centre of circle to point D is
`tantheta=((3)/(2)-1)/((3sqrt3)/(2)-sqrt3)=(1)/(sqrt3)`
It makes an angle `30^(@)` with X-axis.
`therefore` Points E and F will make angle `150^(@) and -90^(@)` with X-axis.

`therefore` E and F are given by
`(x-sqrt3)/(cos150^(@))=(y-1)/(sin150^(@))=1`
`and (x-sqrt3)/(cos(-90^(@)))=(y-1)/(sin(-90^(@)))=1`
`therefore E=((sqrt3)/(2),(3)/(2)and F=(sqrt3,0)`
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