The angle of intersection of the circles x^2 | vedclass

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(C) The angle of intersection $	heta$ between two circles is given by $\cos 	heta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2}$,where $d$ is the distance b

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$	an^{-1}\left(\frac{9}{19}ight)$

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(C) The angle of intersection $	heta$ between two circles is given by $\cos 	heta = \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2}$,where $d$ is the distance between the centers.For the first circle $x^2 + y^2 - x + y - 8 = 0$,the center $C_1 = (\frac{1}{2}, -\frac{1}{2})$ and radius $r_1 = \sqrt{(\frac{1}{2})^2 + (-\frac{1}{2})^2 - (-8)} = \sqrt{\frac{1}{4} + \frac{1}{4} + 8} = \sqrt{\frac{17}{2}}$.For the second circle $x^2 + y^2 + 2x + 2y - 11 = 0$,the center $C_2 = (-1, -1)$ and radius $r_2 = \sqrt{(-1)^2 + (-1)^2 - (-11)} = \sqrt{1 + 1 + 11} = \sqrt{13}$.The distance $d$ between centers $C_1$ and $C_2$ is $d = \sqrt{(\frac{1}{2} - (-1))^2 + (-\frac{1}{2} - (-1))^2} = \sqrt{(\frac{3}{2})^2 + (\frac{1}{2})^2} = \sqrt{\frac{9}{4} + \frac{1}{4}} = \sqrt{\frac{10}{4}} = \sqrt{\frac{5}{2}}$.Now,$\cos 	heta = \frac{\frac{17}{2} + 13 - \frac{5}{2}}{2 	imes \sqrt{\frac{17}{2}} 	imes \sqrt{13}} = \frac{6 + 13}{2 	imes \sqrt{\frac{221}{2}}} = \frac{19}{2 	imes \frac{\sqrt{221}}{\sqrt{2}}} = \frac{19}{\sqrt{2} 	imes \sqrt{221}} = \frac{19}{\sqrt{442}}$.Since $\cos 	heta = \frac{19}{\sqrt{442}}$,we have $\sin^2 	heta = 1 - \cos^2 	heta = 1 - \frac{361}{442} = \frac{81}{442}$,so $\sin 	heta = \frac{9}{\sqrt{442}}$.Thus,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{9/\sqrt{442}}{19/\sqrt{442}} = \frac{9}{19}$.Therefore,$	heta = 	an^{-1}\left(\frac{9}{19}ight)$.

The angle of intersection of the circles $x^2 + y^2 - x + y - 8 = 0$ and $x^2 + y^2 + 2x + 2y - 11 = 0$ is

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