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

When the axes are rotated through an angle $\theta$ about origin in anti-clockwise direction and then translated to the new origin $(2,-2)$, if the transformed equation the equation of $x^2+y^2=4$ is $X^2+Y^2+a X+b Y+c=0$ then $a+b+c=$

From a point $P(-4,0)$, two tangents are drawn to the circle $x^2+y^2-4 x-6 y-12=0$ touching the circle at $A$ and $B$. If the equation of the circle passing through $P, A$ and $B$ is $x^2+y^2+2 g x+2 f y+c=0$, then $(g, f)=$