Question

The angle between the circles $x^2 + y^2 + 4x + 2y + 1 = 0$ and $x^2 + y^2 - 2x + 6y - 6 = 0$ is

• A. $ \frac{\pi}{6}$
• B. $ \frac{\pi}{3}$
• C. $ \frac{\pi}{2}$
• D. $\cos^{-1} \frac{7}{16}$

Answer 1

Answer: (D)

Let $\theta$ be the angle of intersection of two given circles having centres $(-g, -f ) = (-2, -1)$ and $(-g_1, -f_1) = (1, -3)$ respectively, Then
$\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| $
$= \left|\frac{-2\left(-2\right) -2\left(3\right)+1-6}{2\times\sqrt{4} \times \sqrt{16}}\right| =\left|\frac{4-6-5}{16}\right| = \frac{7}{16}$
$\Rightarrow \theta =\cos^{-1} \left(\frac{7}{16}\right)$