Match the items in List-I with the items in L | vedclass

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(A-IV, B-I, C-III, D-II) Given circles are $S_\alpha: x^2+y^2+2\alpha x+k=0$ and $S_\beta: x^2+y^2+2\beta y-k=0$,with $k>0$.$(A)$ For $S_\alpha=0$,the

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(A-IV, B-I, C-III, D-II) Given circles are $S_\alpha: x^2+y^2+2\alpha x+k=0$ and $S_\beta: x^2+y^2+2\beta y-k=0$,with $k>0$.$(A)$ For $S_\alpha=0$,the radius is $r = \sqrt{\alpha^2-k}$. For a point circle,$r=0$,so $\alpha^2-k=0 \Rightarrow \alpha = \pm \sqrt{k}$. The center is $(-\alpha, 0) = (\mp \sqrt{k}, 0)$. Thus,the point circles are $(\pm \sqrt{k}, 0)$. Matches (iv).$(B)$ For $S_\beta=0$,the radius is $r = \sqrt{\beta^2-(-k)} = \sqrt{\beta^2+k}$. Since $k>0$,$\beta^2+k > 0$ for all real $\beta$. Thus,$r$ can never be $0$. Point circles do not exist. Matches $(i)$.$(C)$ For $S_\alpha=0$,the radius is $r = \sqrt{\alpha^2-k}$. If $\alpha^2  k$,the circles are real and non-intersecting (as they are a family of circles with centers on the $x$-axis and varying radii). Given the context of such problems,they are non-intersecting. Matches (iii).$(D)$ For $S_\beta=0$,the radius is $r = \sqrt{\beta^2+k}$. These are a family of circles with centers on the $y$-axis. Since they have different radii for different $\beta$,they are intersecting. Matches (ii).Therefore,the correct match is $A-(iv), B-(i), C-(iii), D-(ii)$.

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

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