If `(x+y)^2=2(x^2+y^2)` and `(x-y+lambda)^2=4`, `lambda>0` then `lambda` is equal to

If (x+y)^(2)=2(x^(2)+y^(2)) and (xy+lambda)^(2)=4,lambda>0, then lambda is equal to

The equation of the circle which touches the axes of co-ordination and the line ( x)/( 3) +( y )/( 4) =1 and when centre lines in the first quadrant is x^(2)+y^(2) - 2 lambda x - 2lambda y +lambda^(2) =0 , then lambda is equal to :

If the area (in sq.units) bounded by the parabola y^(2)=4 lambda x and the line y=lambda x,lambda>0, is (1)/(9), then lambda is equal to:

If x=(4 lambda)/(1+lambda^(2)) and y=(2-2 lambda^(2))/(1+lambda^(2)) where lambda is a real parameter and Z=x^(2)+y^(2)-xy then possible values of Z can be

If x+y+1 = 0 touches the parabola y^(2) = lambda x , then lambda is equal to

The area between the curves x=4y-y^(2) and 0 is lambda square units, then the value of 3lambda is equal to

The area between the curves x=4y-y^(2) and 0 is lambda square units, then the value of 3lambda is equal to

2x^2 +2y^2 +4 lambda x+lambda =0 represents a circle for