`x^2-4x+4=0`

Solve the following quadratic equation: \ i x^2-4x-4i=0

A quadratic equation whose roots are cosec^2theta and sec^2theta can be (1) x^2-2x+ 2 =0 (2) x^2-3x + 3 = 0 (3) x^2-3x+ 4= 0 (4) x^2-5x+5= 0

Remove the second term from the equation x^4 +4x^3 +2x^2 -4x -2=0

Solve the following quadratic equation: \ x^2+4i x-4=0

Find the transformed equation of 2x^(2)+y^(2)-4x+4y=0 when the origin is shifted to the point (-1,2).

The coordinates of the point where origin is shifted is (-1,2) so that the equation 2x^(2)+y^(2)-4x+4y=0 become?

If the points A(2, 5) and B are symmetrical about the tangent to the circle x^(2)+y^(2)-4x+4y=0 at the origin, then the coordinates of B, are

a circle passing through the point (2,2(sqrt2-1)) touches the pair of lines x^2 -y^2 -4x+4 =0 . The centre of the circle is

The locus of the intersection points of pair of perpendicular tangents to the parabola x^2-4x + 4y-8=0 . is

If the line a x+b y=2 is a normal to the circle x^2+y^2-4x-4y=0 and a tangent to the circle x^2+y^2=1 , then