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=0 is

The equation of circle which pass through the points (2.0),(0.2) and orthogonal to the circle 2x^(2)+2y^(2)+5x-6y+4=0

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.

Tangents drawn from P(1,8) to the circle x^(2)+y^(2)-6x-4y-11=0 touches the circle at the points A and B, respectively.The radius of the circle which passes through the points of intersection of circles x^(2)+y^(2)-2x-6y+6=0 and x^(2)+y^(2)-2x-6y+6=0 the circumcircle of the and interse Delta PAB orthogonally is equal to