The radius of the circle which cuts all the three circles $x^2+y^2-4x-4y+3=0$,$x^2+y^2+4x-4y+3=0$,and $x^2+y^2+4x+4y+3=0$ orthogonally is

If the circles $x^2+y^2-2x-2(3+\sqrt{7})y+8+6\sqrt{7}=0$ and $x^2+y^2-8x-6y+k^2=0, k \in \mathbb{Z}$,have exactly two common tangents,then the number of possible values of $k$ 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

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 $k$ is equal to :

If the two circles $2x^2 + 2y^2 - 3x + 6y + k = 0$ and $x^2 + y^2 - 4x + 10y + 16 = 0$ intersect orthogonally,then the value of $k$ is:

$A$ circle $S$ cuts three circles $x^2+y^2-4x-2y+4=0$,$x^2+y^2-2x-4y+1=0$,and $x^2+y^2+4x+2y+1=0$ orthogonally. Then,the radius of $S$ is