Question

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

• A. $tan^{-1} \left(\frac{19}{9}\right)$
• B. $tan^{-1} \left(19\right)$
• C. $tan^{-1} \left(\frac{9}{19}\right)$
• D. $tan^{-1} \left(9\right)$

Answer 1

Answer: (C)

$tan^{-1} \left(\frac{9}{19}\right)$

Answer 2

We know that, angle of intersection between two circles is given by
$\cos \theta=\frac{r_{1}^{2}+r_{2}^{2}-d^{2}}{2 r_{1} r_{2}}=\frac{\frac{17}{2}+13-\frac{10}{4}}{2 \sqrt{\frac{17}{2}} \cdot \sqrt{13}}$

$\Rightarrow \cos \theta=\left(\frac{19}{\sqrt{442}}\right)$
or $\tan \theta=\left(\frac{9}{19}\right)$
$\Rightarrow \theta=\tan ^{-1}\left(\frac{9}{19}\right)$