lambda x^(2)+(2 lambda-3)y^(2)-4x+6y-1=0" र्छिडि "6" ."

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

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

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

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 (4 lambda - 3) x^(2) + lambda y ^(2) + 6x - 2y + 2 = 0 represents a circle, then its centre is

If 3x^(2)+2lambda xy + 3y^(2)+(6-lambda)x+(2lambda-6)y-21=0 is the equation of a circle, then its radius is

If 3x^(2)+2lambda xy + 3y^(2)+(6-lambda)x+(2lambda-6)y-21=0 is the equation of a circle, then its radius is

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

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