If the acute angle between the circles S equi | vedclass

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(A) The circles are $S: x^2+y^2+2kx+4y-3=0$ and $S': x^2+y^2-4x+2ky+9=0$. The centres are $C_1 = (-k, -2)$ and $C_2 = (2, -k)$. The radii are $r_1 = \

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$x-5y+6=0$

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(A) The circles are $S: x^2+y^2+2kx+4y-3=0$ and $S': x^2+y^2-4x+2ky+9=0$. The centres are $C_1 = (-k, -2)$ and $C_2 = (2, -k)$. The radii are $r_1 = \sqrt{k^2+4+3} = \sqrt{k^2+7}$ and $r_2 = \sqrt{4+k^2-9} = \sqrt{k^2-5}$. The distance between centres is $d^2 = (2+k)^2 + (-k+2)^2 = 4+4k+k^2 + k^2-4k+4 = 2k^2+8$. The angle $	heta$ between circles satisfies $\cos 	heta = \frac{d^2-r_1^2-r_2^2}{2r_1r_2}$. Given $\cos 	heta = \frac{3}{8}$,so $\frac{2k^2+8-(k^2+7)-(k^2-5)}{2\sqrt{k^2+7}\sqrt{k^2-5}} = \frac{3}{8}$. $\frac{6}{2\sqrt{(k^2+7)(k^2-5)}} = \frac{3}{8} \implies \frac{3}{\sqrt{k^4+2k^2-35}} = \frac{3}{8}$. $\sqrt{k^4+2k^2-35} = 8 \implies k^4+2k^2-35 = 64 \implies k^4+2k^2-99 = 0$. $(k^2+11)(k^2-9) = 0$. Since $k^2 > 0$,$k^2=9$,so $k=3$ or $k=-3$. The centre of $S'=0$ is $(2, -k)$. For it to be in the first quadrant,$-k > 0$,so $k=-3$. The radical axis is $S-S'=0$: $(2k+4)x + (4-2k)y - 12 = 0$. Substituting $k=-3$: $(2(-3)+4)x + (4-2(-3))y - 12 = 0 \implies -2x + 10y - 12 = 0 \implies x-5y+6=0$.

List-$I$	List-$II$
$(A)$ $(x-2)^2+y^2=2$,$(x-2)^2+(y-1)^2=1$	$I.$ $90^{\circ}$
$(B)$ $x^2+y^2-6x-6y+9=0$,$x^2+y^2-4x+4y-9=0$	$II.$ $135^{\circ}$
$(C)$ $x^2+y^2+4x-14y+28=0$,$x^2+y^2+4x-5=0$	$III.$ $60^{\circ}$
	$IV.$ $30^{\circ}$

If the acute angle between the circles $S \equiv x^2+y^2+2kx+4y-3=0$ and $S' \equiv x^2+y^2-4x+2ky+9=0$ is $\cos^{-1}(\frac{3}{8})$ and the centre of $S'=0$ lies in the first quadrant,then the radical axis of $S=0$ and $S'=0$ is

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