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

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 :

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

The equations of two circles are x^(2)+y^(2)+2 lambda x+5=0 and x^(2)+y^(2)+2 lambda y+5=0.P is any point on the line x-y=0. If PA and PB are the lengths of the tangent from Ptothe circles and PA=3 then find PB.

The value of the lambda if the lines (2+3y+4)+lambda(6x-y+12)=0 are