Find the centre and the radius of the circles
`2x^(2) + lambda xy + 2y^(2) + (lambda - 4) x+6y - 5 = 0`, for some `lambda`

The coordinates of the centre and radius of the circle represented by the equation (3-2lambda)x^(2)+lambda y^(2)-4x+2y-4=0 are

The coordinates of the centre and radius of the circle represented by the equation (3-2lambda)x^(2)+lambda y^(2)-4x+2y-4=0 are

Coordinates of the centre of the circle given by the equation lambda x^(2) + (2lambda - 3)y^(2) - 4x + 6y - 1 =0 is ………

If the equation of a circle is lambda x^(2) + (2 lambda - 3)y^(2) - 4x + 6y - 1 = 0 , then the coordinates of centre are -

The centre of the circle lambdax^2+(2lambda-3)y^2-4x+6y-1=0 is

If 2x-3y=0 is the equation of the common chord of the circles x^(2)+y^(2)+4x=0 and x^(2)+y^(2)+2 lambda y=0 then the value of lambda is

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 :

The line 4y - 3x + lambda =0 touches the circle x^2 + y^2 - 4x - 8y - 5 = 0 then lambda=