The radical centre of the circles $x^2+y^2=1$,$x^2+y^2-2x-3=0$ and $x^2+y^2-2y-3=0$ is

The radical centre of three circles described on the three sides of a triangle as diameters is

If two circles touch at $(0,0)$ externally,then their radical axis is

If $(h, k)$ is the centre of the circle which passes through the origin and cuts the circles $x^2+y^2+4x+6y+12=0$ and $x^2+y^2+4x-6y+9=0$ orthogonally,then $k-2h=$

The straight line $x+y-1=0$ meets the circle $x^2+y^2-6x-8y=0$ at $A$ and $B$. Then the equation of the circle of which $AB$ is a diameter 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| = $