Question

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

• 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

The correct option is C ${\tan }^{-1}\left(\frac{9}{19}\right)$

Angle of intersection between two circles is given by
$\cos \theta =\frac{r_1^2+r_2^2-d^2}{2r_1{r}_2}=\frac{\frac{17}{2}+13-\frac{10}{4}}{2\sqrt{\frac{17}{2}}.\sqrt{13}}$
$\{\begin{array}{c}here,r_1=\sqrt{{\left(\frac{1}{2}\right)}^2+{(-\frac{1}{2})}^2+8}\\ =\sqrt{\frac{17}{2}},\\ r_2=\sqrt{13}\\ andd=c_1{c}_2=\sqrt{\frac{10}{2}}\end{array}$
$\Rightarrow \cos \theta =\left(\frac{19}{\sqrt{442}}\right)$
or $\tan \theta =\left(\frac{9}{19}\right)$
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{9}{19}\right)$