The equation of the circle passing through the origin & cutting the circles `x^2 + y^2-4x+ 6y + 10 = 0` and `x^2 + y^2 +12y + 6=0` at right angles is -

Find the equation of circle passing through the origin and cutting the circles x^(2) + y^(2) -4x + 6y + 10 =0 and x^(2) + y^(2) + 12y + 6 =0 orthogonally.

Find the equation of circle passing through the origin and cutting the circles x^(2) + y^(2) -4x + 6y + 10 =0 and x^(2) + y^(2) + 12y + 6 =0 orthogonally.

The circles x^2 + y^2 + 6x + 6y = 0 and x^2 + y^2 - 12x - 12y = 0

The equation of the circle passing through (0,0) and cutting orthogonally the circles x^(2) + y^(2) + 6x - 15 = 0 , x^(2) + y^(2) - 8y + 10 = 0 is

The equation of the circle passing through (1,2) and the points of intersection of the circles x^2+y^2-8x-6y+21=0 and x^2+y^2-2x-15=0 is

The equation of the circle passing through (1,2) and the points of intersection of the circles x^2+y^2-8x-6y+21=0 and x^2+y^2-2x-15=0 is

The equation of the circle which pass through the origin and cuts orthogonally each of the circles x^2+y^2-6x+8=0 and x^2+y^2-2x-2y=7 is

The equation of the circle which pass through the origin and cuts orthogonally each of the circles x^2+y^2-6x+8=0 and x^2+y^2-2x-2y=7 is