Q118 / 180 · COMEDK 2012 · Mathematics
The angle between the circles
$\begin{aligned}
&x^{2}+y^{2}+4 x+2 y+1=0 \text { and } \\
&x^{2}+y^{2}-2 x+6 y-6=0 \text { is }
\end{aligned}$
- A$\frac{\pi}{6}$
- B$\frac{\pi}{3}$
- C$\frac{\pi}{2}$
- Correct answerD$\cos ^{-1}\left(\frac{7}{16}\right)$
Solution
Let ${\theta}$ be the angle of intersection of two given circles having centres $(-g,-f)=(-2,-1)$ and $\left(-g_{1},-f_{1}\right)=(1,-3)$ respectively, then
$$
\begin{aligned}
\cos \theta &=\left|\frac{-2 g g_{1}-2 f f_{1}+c+c_{1}}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g_{1}^{2}+f_{1}^{2}-c_{1}}}\right| \\
&=\left|\frac{-2(2)(-1)-2(1)(3)+1-6}{2 \sqrt{4+1-1} \sqrt{1+9+6}}\right| \\
&=\left|\frac{4-6-5}{2(2)(4)}\right|=\left(\frac{7}{6}\right) \\
\theta &=\cos ^{-1}\left(\frac{7}{16}\right)
\end{aligned}
$$
Hence, the correct answer is option: 'D'.