Consider the three circles: S_{1} equiv x^{2} | vedclass

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(C) The radical centre of three circles exists if and only if the radical axes of the circles are not parallel,i.e.,they are not concurrent at infinit

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$5$

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(C) The radical centre of three circles exists if and only if the radical axes of the circles are not parallel,i.e.,they are not concurrent at infinity. First,find the radical axis of $S_{1}$ and $S_{2}$ by $S_{1} - S_{2} = 0$: $(x^{2}+y^{2}-6x-6y+4) - (x^{2}+y^{2}-2x-4y+3) = 0$ $-4x - 2y + 1 = 0 \Rightarrow 4x + 2y - 1 = 0$. Next,find the radical axis of $S_{2}$ and $S_{3}$ by $S_{2} - S_{3} = 0$: $(x^{2}+y^{2}-2x-4y+3) - (x^{2}+y^{2}+2kx+2y+1) = 0$ $-(2+2k)x - 6y + 2 = 0 \Rightarrow (2+2k)x + 6y - 2 = 0$. The radical centre does not exist if these two lines are parallel. Two lines $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$ are parallel if $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}$. So,$\frac{4}{2+2k} = \frac{2}{6}$. $24 = 2(2+2k)$ $\Rightarrow 24 = 4 + 4k$ $\Rightarrow 20 = 4k$ $\Rightarrow k = 5$. Thus,if $k = 5$,the lines are parallel and the radical centre does not exist. Therefore,$k$ cannot be $5$.

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

Consider the three circles: $S_{1} \equiv x^{2}+y^{2}-6x-6y+4=0$,$S_{2} \equiv x^{2}+y^{2}-2x-4y+3=0$,and $S_{3} \equiv x^{2}+y^{2}+2kx+2y+1=0$. If the radical centre of these three circles exists,then which of the following cannot be the value of $k$?

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