If the line $x+y=2$ cuts the circle $x^2+y^2+2x-4y+4=0$ at two points $A$ and $B$,then the radius of the circle passing through $A$ and $B$ and orthogonal to $x^2+y^2-2x-4y-4=0$ is

$P, Q$ and $R$ are the centres and $r_1, r_2, r_3$ are the radii respectively of three co-axial circles. Then $QRr_1^2 + RP r_2^2 + PQ r_3^2$ is equal to

The equation of the radical axis of the pair of circles $7x^2+7y^2-7x+14y+18=0$ and $4x^2+4y^2-7x+8y+20=0$ is

The lengths of tangents from a fixed point to three circles of a coaxial system are $t_1, t_2, t_3$. If $P, Q,$ and $R$ are the centers of these circles,then $QRt_1^2 + RPt_2^2 + PQt_3^2$ is equal to

If $\left(0, \frac{3}{4}\right)$ is the radical centre of the circles $S \equiv x^2+y^2+\alpha x+6y=0$,$S^{\prime} \equiv x^2+y^2+2\alpha x+\alpha y+6=0$ and $S^{\prime\prime} \equiv x^2+y^2+6\alpha x-\alpha y+3=0$,then the distance between the radical centre and the centre of the circle $S^{\prime}=0$ is:

If the circle $x^2+y^2+8x-4y+c=0$ touches the circle $x^2+y^2+2x+4y-11=0$ externally and cuts the circle $x^2+y^2-6x+8y+k=0$ orthogonally,then $k$ is equal to