If the angle between the circles x^2+y^2-12x- | vedclass

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(B) The angle $	heta$ between two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ is given by $\cos 	heta = \frac{d^2-r_1^2-r_2^

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$-4$

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(B) The angle $	heta$ between two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ is given by $\cos 	heta = \frac{d^2-r_1^2-r_2^2}{2r_1r_2}$,where $d$ is the distance between the centers $(C_1, C_2)$ and $r_1, r_2$ are the radii.For $C_1: x^2+y^2-12x-6y+41=0$,center is $(6, 3)$ and $r_1 = \sqrt{6^2+3^2-41} = \sqrt{36+9-41} = 2$.For $C_2: x^2+y^2+kx+6y-59=0$,center is $(-\frac{k}{2}, -3)$ and $r_2 = \sqrt{(-\frac{k}{2})^2+(-3)^2-(-59)} = \sqrt{\frac{k^2}{4}+68}$.The distance $d^2 = (6+\frac{k}{2})^2 + (3+3)^2 = (6+\frac{k}{2})^2 + 36$.Given $	heta = 45^{\circ}$,so $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$.Substituting values: $\frac{1}{\sqrt{2}} = \frac{(6+\frac{k}{2})^2+36-4-(\frac{k^2}{4}+68)}{2(2)\sqrt{\frac{k^2}{4}+68}} = \frac{36+6k+\frac{k^2}{4}+36-4-\frac{k^2}{4}-68}{4\sqrt{\frac{k^2+272}{4}}} = \frac{6k}{2\sqrt{k^2+272}} = \frac{3k}{\sqrt{k^2+272}}$.Squaring both sides: $\frac{1}{2} = \frac{9k^2}{k^2+272}$ $\Rightarrow k^2+272 = 18k^2$ $\Rightarrow 17k^2 = 272$ $\Rightarrow k^2 = 16$ $\Rightarrow k = \pm 4$.Thus,the value of $k$ is $-4$.

If the angle between the circles $x^2+y^2-12x-6y+41=0$ and $x^2+y^2+kx+6y-59=0$ is $45^{\circ}$,then a value of $k$ is

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