The slope of one of the direct common tangents drawn to the circles $x^2+y^2-2 x+4 y+1=0$ and $x^2+y^2-4 x-2 y+4=0$ is

If the pole of the line $x+2 b y-5=0$ with respect to the circle $S \equiv x^2+y^2-4 x-6 y+4=0$ lies on the line $x+b y+1=0$, then the polar of the point $(b,-b)$ with respect to the circle $S=0$ is

If $P(\alpha, \beta)$ is the radical centre of the circles $S \equiv x^2+y^2+4 x+7=0, S^{\prime}=2 x^2+2 y^2+3 x+5 y+9=0$ and $S^{\prime \prime} \equiv x^2+y^2+y=0$, then the length of the tangent drawn from $P$ to $S^{\prime}=0$ is