If the radical centre of the circles $x^2+y^2-8x-2y+8=0$,$x^2+y^2+6x+8y-24=0$,and $x^2+y^2-2x+2y+2=0$ is $(a, b)$,then $a+b=$

If the equation of the circle of radius $3$ units which touches the circle $x^2+y^2+6x-8y-11=0$ externally at $(3,0)$ is $x^2+y^2+2gx+2fy+c=0$,then $3g-4f+c=$

$A$ tangent $PT$ is drawn to the circle $x^2 + y^2 = 4$ at the point $P(\sqrt{3}, 1)$. $A$ line $L$ perpendicular to $PT$ is a tangent to the circle $(x - 3)^2 + y^2 = 1$. The possible equation of $L$ is:

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) $(\pm \sqrt{k}, 0)$
	$(v)$ $(0, \pm \sqrt{k})$

The radical axis of the circles $x^2+y^2+2gx+2fy+c=0$ and $2x^2+2y^2+3x+8y+2c=0$ touches the circle $x^2+y^2+2x+2y+1=0$. Then

One of the limit points of the coaxial system of circles containing $x^2 + y^2 - 6x - 6y + 4 = 0$ and $x^2 + y^2 - 2x - 4y + 3 = 0$ is: