If the circle x^2+y^2+2kx+4y-4=0 has its cent | vedclass

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(D) For the circle $x^2+y^2+2kx+4y-4=0$,the centre $C_1 = (-k, -2)$ and radius $R_1 = \sqrt{k^2+4+4} = \sqrt{k^2+8}$.For the circle $x^2+y^2+6x-2y+6=0

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(D) For the circle $x^2+y^2+2kx+4y-4=0$,the centre $C_1 = (-k, -2)$ and radius $R_1 = \sqrt{k^2+4+4} = \sqrt{k^2+8}$.For the circle $x^2+y^2+6x-2y+6=0$,the centre $C_2 = (-3, 1)$ and radius $R_2 = \sqrt{9+1-6} = 2$.Since the circles touch each other,the distance between their centres $C_1C_2 = R_1 + R_2$ (assuming external touch).$C_1C_2 = \sqrt{(-3+k)^2 + (1+2)^2} = \sqrt{k^2-6k+9+9} = \sqrt{k^2-6k+18}$.Setting $C_1C_2 = R_1 + R_2$: $\sqrt{k^2-6k+18} = \sqrt{k^2+8} + 2$.Squaring both sides: $k^2-6k+18 = k^2+8 + 4 + 4\sqrt{k^2+8}$.$6-6k = 4\sqrt{k^2+8} \Rightarrow 3(1-k) = 2\sqrt{k^2+8}$.Squaring again: $9(1-2k+k^2) = 4(k^2+8) \Rightarrow 9k^2-18k+9 = 4k^2+32$.$5k^2-18k-23 = 0 \Rightarrow (k+1)(5k-23) = 0$.So,$k = -1$ or $k = \frac{23}{5}$.The centre is $(-k, -2)$. For the $4^{	ext{th}}$ quadrant,the $x$-coordinate must be positive,so $-k > 0$,which means $k Thus,$k = -1$.

If the circle $x^2+y^2+2kx+4y-4=0$ has its centre in the $4^{\text{th}}$ quadrant and touches the circle $x^2+y^2+6x-2y+6=0$,then $k=$

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