If the equation of the circle which cuts each of the circles $x^{2}+y^{2}=4, x^{2}+y^{2}-6 x-8 y+10=0$ and $x^{2}+y^{2}+2 x-4 y-2=0$ at the extremities of a diameter of these circles is $x^{2}+y^{2}+2 g x+2 f y+c=0$, then $g+f+c=$

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

$(1, k)$ is a point on the circle passing through the points $(-1,1),(0,-1)$ and $(1,0)$. If $k \neq 0$, then $k=$

If the tangents $x+y+k=0$ and $x+a y+b=0$ drawn to the circle $S=x^2+y^2+2 x-2 y+1=0$ are perpendicular to each other and $k, b$ are both greater than 1 , then $b-k=$