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

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

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 the circle ${x^2} + {y^2} + 6x - 2y + k = 0$ bisects the circumference of the circle ${x^2} + {y^2} + 2x - 6y - 15 = 0,$ then $k =$

The circle on the chord $x\cos \alpha + y\sin \alpha = p$ of the circle ${x^2} + {y^2} = {a^2}$ as diameter has the equation

Let $C_1$ and $C_2$ be the centres of the circles $x^2 + y^2 -2x -2y -2 = 0$ and $x^2 + y^2 - 6x-6y + 14 = 0$ respectively. If $P$ and $Q$ are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral $PC_1QC_2$ is ............. $\mathrm{sq. \, units}$