If $(h, k)$ is the centre of the circle which passes through the origin and cuts the circles $x^2+y^2+4 x+6 y+12=0$ and $x^2+y^2+4 x-6 y+9=0$ orthogonally, then $k-2 h=$

If $(-1,-1)$ is the radical centre of the circles $x^2+y^2+2 g x-4 y+4=0, x^2+y^2+6 x+2 f y+12=0$ and $x^2+y^2+10 y+20=0$, then $g-f=$

The equation of the given curve is $x^2-4 x+4 y-8=0$. Match the following

$$ \begin{array}{lll} \hline & \text { List I } & \text { List II } \\ \hline \text { (A) } & \text { Focus } & \text { (I) }(4,2) \\ \hline \text { (B) } & \text { Vertex } & \text { (II) }(3,2) \\ \hline \text { (C) } & \begin{array}{l} \text { One end of the } \\ \text { latusrectum } \end{array} & \text { (III) }(2,3) \\ \hline \text { (D) } & \begin{array}{l} \text { point of intersection of the } \\ \text { axis and directrix } \end{array} & \text { (IV) }(2,4) \\ \hline & & \text { (V) }(2,2) \\ \hline \end{array} $$

$$ \text { The correct match is } $$

If one end of a focal chord of the parabola $y^2=\frac{8}{a} \times(a>0)$ is at $(1,4)$, then the length of this focal chord is