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 and D, E, F` respectively. The line `PQ` is given by the equation `sqrt(3) +y-6=0` and the point `D` is `((3sqrt(3))/(2), 3/2)` The equation of circle `C` is : (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`

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 ((3sqrt3)/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. y=(2)/(sqrt3)+x+1,y=-(2)/(sqrt3)x-1 B. y=(1)/(sqrt3)x,y=0 C. y=(sqrt3)/(2)x+1,y=-(sqrt3)/(2)x-1 D. y=sqrt3x,y=0

Find the equation of the normal to the circle x^2+y^2=9 at the point (1/(sqrt(2)),1/(sqrt(2))) .

The line 2x-y+1=0 is tangent to the circle at the point (2, 5) and the center of the circle lies on x-2y=4. The radius of the circle is (a) 3sqrt(5) (b) 5sqrt(3) (c) 2sqrt(5) (d) 5sqrt(2)

An equilateral triangle whose two vertices are (-2, 0) and (2, 0) and which lies in the first and second quadrants only is circumscribed by a circle whose equation is : (A) sqrt(3)x^2 + sqrt(3)y^2 - 4x +4 sqrt(3)y = 0 (B) sqrt(3)x^2 + sqrt(3)y^2 - 4x - 4 sqrt(3)y = 0 (C) sqrt(3)x^2 + sqrt(3)y^2 - 4y + 4 sqrt(3)y = 0 (D) sqrt(3)x^2 + sqrt(3)y^2 - 4y - 4 sqrt(3) = 0

The incenter of the triangle with vertices (1,sqrt(3)),(0,0), and (2,0) is (a) (1,(sqrt(3))/2) (b) (2/3,1/(sqrt(3))) (c) (2/3,(sqrt(3))/2) (d) (1,1/(sqrt(3)))

A solution of sin^-1 (1) -sin^-1 (sqrt(3)/x^2)- pi/6 =0 is (A) x=-sqrt(2) (B) x=sqrt(2) (C) x=2 (D) x= 1/sqrt(2)

Let y=1/(2+1/(3+1/(2+1/(3+)))) , then the value of y is (a) (sqrt(13)+3)/2 (b) (sqrt(13)-3)/2 (c) (sqrt(15)+3)/2 (d) (sqrt(15)-3)/2

The equation of a circle of radius 1 touching the circles x^2+y^2-2|x|=0 is (a) x^2+y^2+2sqrt(2)x+1=0 (b) x^2+y^2-2sqrt(3)y+2=0 (c) x^2+y^2+2sqrt(3)y+2=0 (d) x^2+y^2-2sqrt(2)+1=0

To which of the circles, the line y-x+3=0 is normal at the point (3+3sqrt2, 3sqrt2) is

If the line y=x+sqrt(3) touches the ellipse (x^(2))/(4)+(y^(2))/(1)=1 then the point of contact is