The radical axis of the circles $S_1: x^2+y^2-4x+6y-10=0$ and $S_2: x^2+y^2+2x-6y+2=0$ cuts the circle $S_1$ in

If $\left(0, \frac{3}{4}\right)$ is the radical centre of the circles $S \equiv x^2+y^2+\alpha x+6y=0$,$S^{\prime} \equiv x^2+y^2+2\alpha x+\alpha y+6=0$ and $S^{\prime\prime} \equiv x^2+y^2+6\alpha x-\alpha y+3=0$,then the distance between the radical centre and the centre of the circle $S^{\prime}=0$ is:

The radical axis of the circles $x^2+y^2+5x+4y-5=0$ and $x^2+y^2-3x+5y-6=0$ 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

The centre of the smallest circle which cuts the circles $x^2+y^2-2x-4y-4=0$ and $x^2+y^2-10x+12y+52=0$ orthogonally is

The value of $\lambda$,for which the circle $x^2 + y^2 + 2\lambda x + 6y + 1 = 0$ intersects the circle $x^2 + y^2 + 4x + 2y = 0$ orthogonally is