The centre of the circle, which cuts orthogonally each of the three circles ${x^2} + {y^2} + 2x + 17y + 4 = 0,$ ${x^2} + {y^2} + 7x + 6y + 11 = 0,$ ${x^2} + {y^2} - x + 22y + 3 = 0$ is

$P$ is a point $(a, b)$ in the first quadrant. If the two circles which pass through $P$ and touch both the co-ordinate axes cut at right angles, then :

The circles ${x^2} + {y^2} + 4x + 6y + 3 = 0$ and $2({x^2} + {y^2}) + 6x + 4y + C = 0$ will cut orthogonally, if $C$ equals

The equation of the circle passing through the point $(1, 2)$ and through the points of intersection of $x^2 + y^2 - 4x - 6y - 21 = 0$ and $3x + 4y + 5 = 0$ is given by

Two given circles ${x^2} + {y^2} + ax + by + c = 0$ and ${x^2} + {y^2} + dx + ey + f = 0$ will intersect each other orthogonally, only when

If circles ${x^2} + {y^2} + 2ax + c = 0$and ${x^2} + {y^2} + 2by + c = 0$ touch each other, then