`x^2 -c^2 =y` then num radius of `Delta` is (A) `y/sqrt3` (B) `c/sqrt3` (C) `(3y)/4` (D)` (3c)/4`

if sum of two sides of a triangle is x and the product of those sides is y and third side is c such that x^2 -c^2 =y then circum radius of Delta is (A) y/sqrt3 (B) c/sqrt3 (C) (3y)/4 (D) (3c)/4

x^(2)-c^(2)=y then num radius of Delta is (A)(y)/(sqrt(3)) (B) (c)/(sqrt(3))(C)(3y)/(4)(D)(3c)/(4)

In a Delta ABC if A=(1,2) and the internal angle bisectors through B and C are y=x and y=-2x. The inradius r of Delta ABC is (A) 1/sqrt3 (B) 1/sqrt2 (C) 2/3 (D) none of these

If a^2x^4+b^2y^4=c^6, then the maximum value of x y is (a) (c^2)/(sqrt(a b)) (b) (c^3)/(a b) (c) (c^3)/(sqrt(2a b)) (d) (c^3)/(2a b)

The area of the circle x^2+y^2=16 exterior to the parabola y^2=6x is (A) 4/3(4pi-sqrt(3)) (B) 4/3(4pi+sqrt(3)) (C) 4/3(8pi-sqrt(3)) (D) 4/3(8pi+sqrt(3))

The area of the circle x^2+y^2=16 exterior to the parabola y^2=6x is(A) 4/3(4pi-sqrt(3)) (B) 4/3(4pi+sqrt(3)) (C) 4/3(8pi-sqrt(3)) (D) 4/3(8pi+sqrt(3))

If a^2x^4+b^2y^4=c^6, then the maximum value of x y is (c^2)/(sqrt(a b)) (b) (c^3)/(a b) (c^3)/(sqrt(2a b)) (d) (c^3)/(2a b)

If cosxdy/dx-y sin x=6x, (0ltxltx/2)and y(pi/3)=0 , then y(pi/6) is equal to (a) pi^(2) /(2sqrt 3) (b) -pi^(2) /(2sqrt 3) (c) -pi^(2) /(4sqrt 3) (d) (c) -pi^(2) /2

The shortest distance between the line yx=1 and the curve x=y^(2) is (A)(3sqrt(2))/(8) (B) (2sqrt(3))/(8) (C) (3sqrt(2))/(5) (D) (sqrt(3))/(4)

A vertex of an equilateral triangle is at (2, 3), and th equation of the opposite side is x+y=2 , then the equaiton of the other two sides are (A) y=(2+sqrt(3)) (x-2), y-3=2sqrt(3)(x-2) (B) y-3=(2+sqrt(3) (x-2), y-3= (2-sqrt(3) (x-2) (C) y+3=(2-sqrt(3)(x-2), y-3=(2-sqrt(3) (x+2) (D) none of these