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

The circles $x^2 + y^2 + 2gx + 2fy = 0$ and $x^2 + y^2 + 2g'x + 2f'y = 0$ touch each other externally if:

If $x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0$ is the smallest circle through the points of intersection of $x^2+y^2=a^2$ and $x \cos \alpha+y \sin \alpha=p$,where $0 < p < a$,then $\lambda=$

The slope of one of the direct common tangents drawn to the circles $x^2+y^2-2x+4y+1=0$ and $x^2+y^2-4x-2y+4=0$ is

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

The line $Ax + By + C = 0$ cuts the circle $x^2 + y^2 + ax + by + c = 0$ at points $P$ and $Q$,and the line $A'x + B'y + C' = 0$ cuts the circle $x^2 + y^2 + a'x + b'y + c' = 0$ at points $R$ and $S$. If the four points $P, Q, R,$ and $S$ are concyclic,then $D = \left| {\begin{array}{*{20}{c}}{a - a'}&{b - b'}&{c - c'}\\A&B&C\\{A'}&{B'}&{C'}\end{array}} \right| = $