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

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Question

(c) $\cos 	heta = \frac{{r_1^2 + r_2^2 - {d^2}}}{{2{r_1}{r_2}}}$

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Vedclass · Accepted Answer

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

Vedclass · Answer

(c) $\cos 	heta = \frac{{r_1^2 + r_2^2 - {d^2}}}{{2{r_1}{r_2}}}$

                                                           ($d$ is the distance between the centres)

${r_1} = \sqrt {\frac{1}{4} + \frac{1}{4} + 8} $ = $\sqrt {\frac{{17}}{2}} $

${r_2} = \sqrt {1 + 1 + 11} $=$\sqrt {13} $

$d = \sqrt {{{\left( { - \frac{1}{2} - 1} ight)}^2} + {{\left( {\frac{1}{2} - 1} ight)}^2}} $= $\sqrt {\frac{5}{2}} $

$	herefore $ $\cos 	heta = \frac{{19}}{{\sqrt {442} }}$

$ \Rightarrow 	an 	heta = \frac{{\sqrt {1 - {{\cos }^2}	heta } }}{{\cos 	heta }} \Rightarrow 	an 	heta = \frac{9}{{19}}$

$ \Rightarrow 	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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