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(A) Let the equation of the required circle be $S \equiv x^2+y^2+2gx+2fy+c=0$. Since the circle $S=0$ cuts the given circles at the extremities of the

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$x^2+y^2-4x-6y-4=0$

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(A) Let the equation of the required circle be $S \equiv x^2+y^2+2gx+2fy+c=0$. Since the circle $S=0$ cuts the given circles at the extremities of their diameters,the common chord of $S=0$ and each circle $S_i=0$ must pass through the center of $S_i=0$. For $S_1 \equiv x^2+y^2-4=0$,the center is $(0,0)$. The common chord is $S-S_1=0$,which is $2gx+2fy+c+4=0$. Since it passes through $(0,0)$,we get $c+4=0$,so $c=-4$. For $S_2 \equiv x^2+y^2-6x-8y+10=0$,the center is $(3,4)$. The common chord is $S-S_2=0$,which is $(2g+6)x+(2f+8)y+c-10=0$. Substituting $c=-4$ and the center $(3,4)$,we get $(2g+6)(3)+(2f+8)(4)-14=0$,which simplifies to $6g+18+8f+32-14=0$,or $6g+8f+36=0$,i.e.,$3g+4f+18=0$ $(i)$. For $S_3 \equiv x^2+y^2+2x-4y-2=0$,the center is $(-1,2)$. The common chord is $S-S_3=0$,which is $(2g-2)x+(2f+4)y+c+2=0$. Substituting $c=-4$ and the center $(-1,2)$,we get $(2g-2)(-1)+(2f+4)(2)-2=0$,which simplifies to $-2g+2+4f+8-2=0$,or $-2g+4f+8=0$ (ii). Solving $(i)$ and (ii): From (ii),$g=2f+4$. Substituting into $(i)$,$3(2f+4)+4f+18=0$ $\Rightarrow 6f+12+4f+18=0$ $\Rightarrow 10f+30=0$ $\Rightarrow f=-3$. Then $g=2(-3)+4=-2$. Thus,the equation is $x^2+y^2-4x-6y-4=0$.

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 equation of the circle which cuts the circles $S_1 \equiv x^2+y^2-4=0$,$S_2 \equiv x^2+y^2-6x-8y+10=0$,and $S_3 \equiv x^2+y^2+2x-4y-2=0$ at the extremities of the diameters of these circles is:

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