The centre of family of circles cutting the family of circles `x^2 + y^2 + 4x(lambda-3/2) + 3y(lambda-4/3)-6(lambda+2)=0` orthogonally, lies on

The differential equation of family of coaxal circles x^(2) + y^(2) + 2lambda x + c = 0 is

The equation to the line joining the centres of the circles belonging to the coaxial system of circles 4x^(2) + 4y^(2) - 12x + 6y - 3 + lambda (x + 2y - 6) = 0 is

The locus of the centre of the circle which cuts the circles x^(2) + y^(2) + 4x - 6y + 9 = 0 " and " x^(2) + y^(2) - 4x + 6y + 4 = 0 orthogonally is

The number of circles belonging to the system of circles 2(x^(2)+y^(2))+lambda x-(1+lambda^(2))y-10=0 and orthogonal to x^(2)+y^(2)+4x+6y+3=0 , is

The number of circles belonging to the system of circles 2(x^(2)+y^(2))+lambda x-(1+lambda^(2))y-10=0 and orthogonal to x^(2)+y^(2)+4x+6y+3=0 , is

Radical axis of any two circles of the family of circle x^(2)+y^(2)+2lambdax+4=0,lambda in R is always parallel to

The locus of the centre of circle which cuts the circles x^2 + y^2 + 4x - 6y + 9 =0 and x^2 + y^2 - 4x + 6y + 4 =0 orthogonally is

Prove that the family of curves x^2/(a^2+lambda)+y^2/(b^2+lambda)=1 , where lambda is a parameter, is self orthogonal.