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 -

The equation of the circle passing through the origin delta 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.

The equation of the circle passing through the points of intersection of the circles x^(2)+y^(2)+6x+4y-12=0 , x^(2)+y^(2)-4x-6y-12=0 and having radius sqrt(13) is

The equation of the circle passing through the points of intersection of the circles x^(2)+y^(2)+6x+4y-12=0 , x^(2)+y^(2)-4x-6y-12=0 and having radius sqrt(13) is

Find the equation of the circle passing through the points of intersection of the circles x^(2)+y^(2)-2x-4y-4=0 and x^(2)+y^(2)-10x-12y+40=0 and whose radius is 4.

Find the centre of the circle that passes through the point (1,0) and cutting the circles x^(2) + y ^(2) -2x + 4y + 1=0 and x ^(2) + y ^(2) + 6x -2y + 1=0 orthogonally 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=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