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. ${\cos }^{-1}\left(\frac{7}{16}\right)$

Answer 1

The correct option is D ${\cos }^{-1}\left(\frac{7}{16}\right)$

Using cosine rule in $△APB$

$A{B}^2=r_1^2+r_2^2-2r_1{r}_2\cos \theta .....\left(1\right)$

Given $x^2+y^2+4x+2y+1=0$ and $x^2+y^2-2x+6y-6=0$

Here centers are $A\equiv (-2,-1),B\equiv (1,-3)$

$r_1=\sqrt{4+1-1}=2,r_2=\sqrt{1+9+6}=4$

$AB=\sqrt{9+4}=\sqrt{13}$

$\left(1\right)\Rightarrow 13=4+16-2\left(2\right)\left(4\right)\cos \theta$

$\Rightarrow 13=20-16\cos \theta$

$\Rightarrow -7=-16\cos \theta$

$\Rightarrow \cos \theta =\frac{7}{16}\Rightarrow \theta ={\cos }^{-1}\frac{7}{16}$