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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
${\tan ^{ - 1}}\left( {\frac{{19}}{9}} \right)$
${\tan ^{ - 1}}(19)$
${\tan ^{ - 1}}\left( {\frac{9}{{19}}} \right)$
${\tan ^{ - 1}}(9)$
Solution
(c) $\cos \theta = \frac{{r_1^2 + r_2^2 – {d^2}}}{{2{r_1}{r_2}}}$
($d$ 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} \right)}^2} + {{\left( {\frac{1}{2} – 1} \right)}^2}} $= $\sqrt {\frac{5}{2}} $
$\therefore $ $\cos \theta = \frac{{19}}{{\sqrt {442} }}$
$ \Rightarrow \tan \theta = \frac{{\sqrt {1 – {{\cos }^2}\theta } }}{{\cos \theta }} \Rightarrow \tan \theta = \frac{9}{{19}}$
$ \Rightarrow \theta = {\tan ^{ – 1}}\left( {\frac{9}{{19}}} \right)$.
