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

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

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

Find the parametric equation of the circles : 3x^2 + 3y^2+4x-6y-4=0 .

If equation of a circle is (4a - 3) x^(2) +ay^(2) + 6x - 2y + 2 = 0 , then its centre is

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

Find the equation of the tangent to the circle x^2+y^2-4x-6y-12=0 at (-1,-1)

Find the parametric equation of the circles : 3x^2 + 3y^2 + 4x-6y - 4 = 0

Find the parametric equation of the circles : 3x^2 + 3y^2 + 4x-6y - 4 = 0

The equation of the normal at (1,2) to the circle x^(2)-y^(2)-4x-6y-12=0 is