The angle between circles $x^2+y^2+2 x+4 y+1=0$ and $x^2+y^2-2 x+6 y-3=0$ is
The angle between circles $x^2+y^2+2 x+4 y+1=0$ and $x^2+y^2-2 x+6 y-3=0$ is
- $\cos ^{-1}\left(\frac{3}{\sqrt{13}}\right)$
- $\cos ^{-1}\left(\frac{3}{\sqrt{31}}\right)$
- $\cos ^{-1}\left(\sqrt{\frac{3}{31}}\right)$
- $2 \cos ^{-1}\left(\frac{3}{\sqrt{13}}\right)$
Solution
Circles are
$
\begin{aligned}
& c_1, x^2+y^2+2 x+4 y+1=0 \\
& c_2, x^2+y^2-2 x+6 y-3=0
\end{aligned}
$
On comparing with general form of circles
$x^2+y^2+2 g x+2 f y+c=0$
where, centre = (− g, − f) and radius
$\begin{gathered}=\sqrt{g^2+f^2-c} \\ \therefore \quad c_1=(-1,-2) \text { and } r_1=\sqrt{1+4-1}=2 \\ \quad c_2=(1,-3) \text { and } r_2=\sqrt{1+9+3}=\sqrt{13} \\ \therefore \quad d=c_1 c_2=\sqrt{(-1-1)^2+(-2+3)^2}=\sqrt{4+1}=\sqrt{5} \\ \therefore \cos \theta=\left|\frac{d^2-r_1^2-r_2^2}{2 r_1 r_2}\right|=\left|\frac{5-4-13}{2 \times 2 \times \sqrt{13}}\right|=\left|\frac{-3}{\sqrt{13}}\right| \\ \Rightarrow \cos \theta=\frac{3}{\sqrt{13}} \\ \therefore \quad \theta=\cos ^{-1}\left(\frac{3}{\sqrt{13}}\right)\end{gathered}$
Asked in: AP EAMCET 2021 (25 Aug Shift 2)
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