If the two circles $2x^2 + 2y^2 - 3x + 6y + k = 0$ and $x^2 + y^2 - 4x + 10y + 16 = 0$ intersect orthogonally,then the value of $k$ is:

$A$ circle $S$ given by $x^2+y^2-14x+6y+33=0$ cuts the $X$-axis at $A$ and $B$ $(OB > OA)$. $C$ is the midpoint of $AB$. $L$ is a line through $C$ with slope $-1$. If $L$ is the diameter of a circle $S^{\prime}$ and also the radical axis of the circles $S$ and $S^{\prime}$,then the equation of the circle $S^{\prime}$ is

$(a, 0)$ and $(b, 0)$ are the centres of two circles belonging to a coaxial system of which the $y$-axis is the radical axis. If the radius of one of the circles is $r$,then the radius of the other circle is

The radical centre of the three circles $x^2+y^2-1=0$,$x^2+y^2-8x+15=0$ and $x^2+y^2+10y+24=0$ is

In List-$I$,a pair of circles is given in $A$,$B$,$C$ and in List-$II$,the angle between those pairs of circles is given. Match the items from List-$I$ to List-$II$.

List-$I$	List-$II$
$(A)$ $(x-2)^2+y^2=2$,$(x-2)^2+(y-1)^2=1$	$I.$ $90^{\circ}$
$(B)$ $x^2+y^2-6x-6y+9=0$,$x^2+y^2-4x+4y-9=0$	$II.$ $135^{\circ}$
$(C)$ $x^2+y^2+4x-14y+28=0$,$x^2+y^2+4x-5=0$	$III.$ $60^{\circ}$
	$IV.$ $30^{\circ}$

The correct matching is

If the circle $x^2 + y^2 + 2x - 4y - k = 0$ lies exactly between the circles $x^2 + y^2 + 2x - 4y - 4 = 0$ and $x^2 + y^2 + 2x - 4y - 20 = 0$,then $k = \dots$