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(B) The given circle is $x^2+y^2+6x+4y-12=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=3, f=2, c=-12$. The radius $r = \sqrt{g^2+f^2-c} = \

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$2 \operatorname{Sin}^{-1}\left(\frac{3}{5}\right)$

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(B) The given circle is $x^2+y^2+6x+4y-12=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=3, f=2, c=-12$. The radius $r = \sqrt{g^2+f^2-c} = \sqrt{9+4+12} = \sqrt{25} = 5$. Let the distance between the centers of the two circles be $d$. The length of the common chord is $L = 2\sqrt{r^2 - (d/2)^2} = 2\sqrt{17}$. Thus,$\sqrt{25 - d^2/4} = \sqrt{17} \implies 25 - d^2/4 = 17 \implies d^2/4 = 8 \implies d^2 = 32 \implies d = 4\sqrt{2}$. Let $	heta$ be the angle between the circles. The formula for the angle between two circles of equal radius $r$ is $\cos 	heta = 1 - \frac{d^2}{2r^2}$. Substituting the values,$\cos 	heta = 1 - \frac{32}{2(25)} = 1 - \frac{32}{50} = 1 - \frac{16}{25} = \frac{9}{25}$. Therefore,$	heta = \operatorname{Cos}^{-1}\left(\frac{9}{25}ight)$. However,the angle between two circles is defined as the angle between their tangents at the point of intersection. For two circles of equal radius,the angle $	heta$ is given by $2 \sin(	heta/2) = L/r$. $2 \sin(	heta/2) = \frac{2\sqrt{17}}{5} \implies \sin(	heta/2) = \frac{\sqrt{17}}{5}$. Using $\cos 	heta = 1 - 2\sin^2(	heta/2) = 1 - 2(17/25) = 1 - 34/25 = -9/25$. Since we need the acute angle,we consider the geometry where $\cos 	heta = |1 - d^2/(2r^2)| = 7/25$ is not correct,the standard formula yields $	heta = 2 \operatorname{Sin}^{-1}(\frac{d}{2r}) = 2 \operatorname{Sin}^{-1}(\frac{4\sqrt{2}}{10}) = 2 \operatorname{Sin}^{-1}(\frac{2\sqrt{2}}{5})$. Re-evaluating,the angle $	heta$ between circles is $2 \operatorname{Sin}^{-1}(\frac{L}{2r}) = 2 \operatorname{Sin}^{-1}(\frac{\sqrt{17}}{5})$.

Length of the common chord of two circles of same radius is $2 \sqrt{17}$. If one of the two circles is $x^2+y^2+6x+4y-12=0$,then the acute angle between the two circles is

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