Angle between the circles $x^2+y^2-4 x-6 y-3=0$ and $x^2+y^2+8 x-4 y+11=0$ is

Angle between the circles $x^2+y^2-4 x-6 y-3=0$ and $x^2+y^2+8 x-4 y+11=0$ is
  1. $\frac{\pi}{3}$
  2. $\frac{\pi}{6}$
  3. $\frac{\pi}{2}$
  4. $\frac{\pi}{4}$

Solution

$\begin{aligned} & \quad \mathrm{S}_1 \equiv x^2+y^2-4 x-6 y-3=0 \\ & \mathrm{~S}_2 \equiv x^2+y^2+8 x-4 y-11=0 \\ & \left(g_1, f_1, c_1\right)=(-2,-3,-3) \text { and }\left(g_2, f_2, c_2\right)=(4,-2,11) \end{aligned}$
Angle between two circles is given by $\begin{aligned} & \cos \theta=\frac{\left|2 g_1 g_2+2 f_1 f_2-c_1-c_2\right|}{2 \sqrt{\left(g_1^2+f_1^2-c_1\right)\left(g_2^2+f_2^2-c_2\right)}} \\ & \Rightarrow \cos \theta=\frac{|-16+12+3-11|}{2 \sqrt{16 \times 9}}=\frac{|-12|}{2 \sqrt{144}}=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 2)

Practice more Circle questions on Aicharya