Consider two curves `C_1 : ( y - sqrt3)^2 = 4 (x- sqrt2) and C_2 : x^2+y^2 = (6+2sqrt2)x + 2 sqrt3 y - 6(1+sqrt2)`, then

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

If x= (sqrt3 - sqrt2)/(sqrt3+sqrt2) and y = (sqrt3+sqrt2)/(sqrt3-sqrt2) then x^2 +xy +y^2 is a multiple of

If x= (sqrt3+sqrt2)/(sqrt3-sqrt2) and y=(sqrt3-sqrt2)/(sqrt3+sqrt2) , then find the value of x^2+y^2

The equation of normal to the curve y=2cosx at x=pi/4 is: a) y-sqrt2=2sqrt2(x-pi/4) b) y+sqrt2=2sqrt2(x+pi/4) c) y-sqrt2=1/sqrt2(x-pi/4) d) y-sqrt2=sqrt2(x-pi/4)