If the circle passing through $(3,5), (5,5)$ and $(3,-3)$ cuts the circle $x^2+y^2+2x+2fy=0$ orthogonally, then $f=$

The combined equation of the direct common tangents of the circles $x^2+y^2+2x=0$ and $x^2+y^2-2y-3=0$ is

Let the circle $S \equiv x^2+y^2+2gx+2fy+c=0$ cut the circles $x^2+y^2-2x+2y-2=0$ and $x^2+y^2+4x-6y+9=0$ orthogonally. If the centre of the circle $S=0$ lies on the line $2x+3y-2=0$,then $2g+f=$

Match the items in List-$I$ with the items in List-$II$ for the circles $S_\alpha: x^2+y^2+2\alpha x+k=0$ and $S_\beta: x^2+y^2+2\beta y-k=0$,where $k>0$.

List-$I$	List-$II$
$(A)$ Point circles of $S_\alpha=0$	$(i)$ do not exist
$(B)$ Point circles of $S_\beta=0$	(ii) intersecting
$(C)$ The circles in $S_\alpha=0$ are	(iii) non-intersecting
$(D)$ The circles in $S_\beta=0$ are	(iv) $(\pm \sqrt{k}, 0)$
	$(v)$ $(0, \pm \sqrt{k})$

If $d$ is the distance between the centres of two circles,$r_1$ and $r_2$ are their radii,and $d = r_1 + r_2$,then

If the circles $(x-2)^2+(y-3)^2=25$ and $25x^2+25y^2-40x-70y-160=0$ touch internally at $(\alpha, \beta)$,then $\alpha+\beta=$