If the circle $S=0$ cuts the circles $x^2+y^2-2 x+6 y=0$, $x^2+y^2-4 x-2 y+6=0$ and $x^2+y^2-12 x+2 y+3=0$ orthogonally, then equation of the tangent at $(0,3)$ on $S=0$ is

If $\theta$ is the angle between the tangents drawn from the point $(2,3)$ to the circle $x^2+y^2-6 x+4 y+12=0$ then $\theta=$

If $2 x-3 y+3=0$ and $x+2 y+k=0$ are conjugate lines with respect to the circle $S=x^2+y^2+8 x-6 y-24=0$, then the length of the tangent drawn from the point $\left(\frac{k}{4}, \frac{k}{3}\right)$ to the circle $S=0$, is

If $Q(h, k)$ is the inverse point of the point $P(1,2)$ with respect to the circle $x^2+y^2-4 x+1=0$, then $2 h+k=$