In the co-axial system of circles $x^2 + y^2 + 2gx + c = 0$,where $g$ is a parameter,if $c > 0$,then the circles are

If $P$ is the point of contact of the circles $x^2+y^2+4x+4y-10=0$ and $x^2+y^2-6x-6y+10=0$,and $Q$ is their external centre of similitude,then the equation of the circle with $P$ and $Q$ as the extremities of its diameter is

The equation of the circle which cuts all the three circles $4(x-1)^2+4(y-1)^2=1$,$4(x+1)^2+4(y-1)^2=1$,and $4(x+1)^2+4(y+1)^2=1$ orthogonally is

$A$ circle $S = x^2 + y^2 + 2gx + 2fy + 4 = 0$ cuts the circle $x^2 + y^2 - 4x - 4y - 4 = 0$ orthogonally and makes an angle of $60^{\circ}$ with the circle $x^2 + y^2 + 4x + 4y + 4 = 0$. Then the radius of the circle $S = 0$ is

The centre of the circle which intersects the circle $x^2+y^2-2x-2y-2=0$ orthogonally,passes through the point $(2,0)$,and touches the $X$-axis is:

If $(h, k)$ is the internal centre of similitude of the circles $x^2+y^2+2x-6y+1=0$ and $x^2+y^2-4x+2y+4=0$,then $4h=$