The image of the centre of the circle `x^2 + y^2 = 2a^2` with respect to the line `x + y = 1` is : (A) `(sqrt(2), sqrt(2)` (B) `(1/sqrt(2) , sqrt(2)`) (C) `(sqrt(2), 1/sqrt(2))` (D) none of these

A(1/sqrt(2), 1/sqrt(2)) is a point on the circle x^2 + y^2 = 1 and B is another point on the circle such that AB = pi/2 units. Then coordinates of B can be : (A) (1/sqrt(2), - 1/sqrt(2)) (B) (-1/sqrt(2), 1/sqrt(2)) (C) (-1/sqrt(2), - 1/sqrt(2)) (D) none of these

A((1)/(sqrt(2)),(1)/(sqrt(2))) is a point on the circle x^(2)+y^(2)=1 and B is another point on the circle such that are length AB=(pi)/(2) units. Then,the coordinates of B can be ((1)/(sqrt(2)),1sqrt(2))( b) (-(1)/(sqrt(2)),1sqrt(2))(-(1)/(sqrt(2)),-(1)/(sqrt(2)))( d) none of these

The point,at shortest distance from the line x+y=10 and lying on the ellipse x^(2)+2y^(2)=6, has coordinates (A) (sqrt(2),sqrt(2)) (B) (0,sqrt(3))(C)(2,1)(D)(sqrt(5),(1)/(sqrt(2)))

Solution set of inequation ( cos^-1x)^2-(sin^-1x)^2gt0 is (A) [0, 1/sqrt(2)) (B) [-1, 1/sqrt(2)) (C) (-1,sqrt2) (D) none of these

(2)/(sqrt(x))+(3)/(sqrt(y))=2 and (4)/(sqrt(x))-(9)/(sqrt(y))=-1

If sqrt(x) + sqrt(y) = sqrt(a) , then (dy)/(dx) = 1/(2sqrt(x)) + 1/(2sqrt(y)) = 1/(2sqrt(a))

The value of lim_(x rarr2)(sqrt(1+sqrt(2+x))-sqrt(3))/(x-2) is (1)/(8sqrt(3))(b)(1)/(4sqrt(3)) (c) 0 (d) none of these

A square is inscribed in the circle x^(2)+y^(2)-2x+4y+3=0. Its sides are parallel to the coordinate axes.One vertex of the square is (1+sqrt(2),-2) (b) (1-sqrt(2),-2)(1,-2+sqrt(2))(d) none of these

{:((2)/(sqrt(x))+ (3)/(sqrt(y)) = 2),((4)/(sqrt(x))-(9)/(sqrt(y)) = -1):}