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

If the circle $S \equiv x^2+y^2+2 g x+2 f y+c=0$ cuts each of the three circles $x^2+y^2+4 x+4 y+7=0$, $x^2+y^2-4 x+4 y+7=0$ and $x^2+y^2-4 x-4 y+7=0$ orthogonally, then the equation of the tangent drawn at the point $(\sqrt{3}, 2)$ to the circle $S=0$ is

Let a chord $A B$ subtend an angle of $60^{\circ}$ at the centre $C(2,3)$ of a circle $S$. If the equation of $A B$ is $x+y+1=0$, then the equation of the circle $S$ is

Let 6,8 be the $X$ and $Y$-intercepts made by the circle $S \equiv x^2+y^2+2 g x+2 f y+c=0$, respectively. If $g x+f y+1=0$ is a line passing through the point $(1,-1)$, then the radius of the circle $S=0$ is