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 A B=pi/2 units. Then, the 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 arc length A B=pi/2 units. Then, the 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

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 (a) (1+sqrt(2),-2) (b) (1-sqrt(2),-2) (c) (1,-2+sqrt(2)) (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 (a) (1+sqrt(2),-2) (b) (1-sqrt(2),-2) (c) (1,-2+sqrt(2)) (d) none of these

The minimum values of sin x + cos x is a) sqrt2 b) -sqrt2 c) (1)/(sqrt2) d) - (1)/(sqrt2)

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)))

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

y=tan^(-1)((sqrt(1+x^2)+sqrt(1-x^2))/(sqrt(1+x^2)-sqrt(1-x^2)))