The equation of the radical axis of the circles ${x^2} + {y^2} - 3x - 4y + 5 = 0$ and $2{x^2} + 2{y^2} - 10x - 12y + 12 = 0$ is:

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:

For the given circles $x^2 + y^2 - 6x - 2y + 1 = 0$ and $x^2 + y^2 + 2x - 8y + 13 = 0$,which of the following is true?

The radical axis of circles $x^2+y^2+5x+4y-5=0$ and $x^2+y^2-3x+5y-6=0$ 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) $(\pm \sqrt{k}, 0)$
	$(v)$ $(0, \pm \sqrt{k})$

If the equation of the circle which cuts each of the circles $x^2+y^2=4$,$x^2+y^2-6x-8y+10=0$,and $x^2+y^2+2x-4y-2=0$ at the extremities of a diameter of these circles is $x^2+y^2+2gx+2fy+c=0$,then $g+f+c=$