If the circles x^2+y^2+2kx-4y+1=0 and x^2+y^2 | vedclass

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(C) For the circle $x^2+y^2+2kx-4y+1=0$,the center $C_1 = (-k, 2)$ and radius $r_1 = \sqrt{(-k)^2 + 2^2 - 1} = \sqrt{k^2+3}$.For the circle $x^2+y^2-8

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(C) For the circle $x^2+y^2+2kx-4y+1=0$,the center $C_1 = (-k, 2)$ and radius $r_1 = \sqrt{(-k)^2 + 2^2 - 1} = \sqrt{k^2+3}$.For the circle $x^2+y^2-8x-12y+43=0$,the center $C_2 = (4, 6)$ and radius $r_2 = \sqrt{4^2 + 6^2 - 43} = \sqrt{16+36-43} = \sqrt{9} = 3$.If two circles touch each other,the distance between their centers is equal to the sum or difference of their radii: $d = |r_1 \pm r_2|$.The distance between centers $C_1(-k, 2)$ and $C_2(4, 6)$ is $d = \sqrt{(4 - (-k))^2 + (6 - 2)^2} = \sqrt{(4+k)^2 + 4^2} = \sqrt{k^2+8k+16+16} = \sqrt{k^2+8k+32}$.Case $1$: $d = r_1 + r_2 \Rightarrow \sqrt{k^2+8k+32} = \sqrt{k^2+3} + 3$.Squaring both sides: $k^2+8k+32 = k^2+3 + 9 + 6\sqrt{k^2+3}$ $\Rightarrow 8k+20 = 6\sqrt{k^2+3}$ $\Rightarrow 4k+10 = 3\sqrt{k^2+3}$.Squaring again: $16k^2+80k+100 = 9(k^2+3) = 9k^2+27 \Rightarrow 7k^2+80k+73 = 0$.Solving for $k$: $k = \frac{-80 \pm \sqrt{6400 - 2044}}{14} = \frac{-80 \pm \sqrt{4356}}{14} = \frac{-80 \pm 66}{14}$.$k = -1$ or $k = -146/14 = -73/7$.Case $2$: $d = |r_1 - r_2| \Rightarrow \sqrt{k^2+8k+32} = |\sqrt{k^2+3} - 3|$.Squaring both sides: $k^2+8k+32 = k^2+3 + 9 - 6\sqrt{k^2+3}$ $\Rightarrow 8k+20 = -6\sqrt{k^2+3}$ $\Rightarrow 4k+10 = -3\sqrt{k^2+3}$.Squaring again: $16k^2+80k+100 = 9(k^2+3) = 9k^2+27 \Rightarrow 7k^2+80k+73 = 0$.This yields the same values for $k$. Checking $k=-1$ in the original equation: $d = \sqrt{1-8+32} = \sqrt{25} = 5$,$r_1 = \sqrt{1+3} = 2$,$r_2 = 3$. Since $d = r_1+r_2$,$k=-1$ is a valid solution.

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 circles $x^2+y^2+2kx-4y+1=0$ and $x^2+y^2-8x-12y+43=0$ touch each other,then $k=$

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