Question

Find the angle of intersection of two circles?

$x^2+y^2+2gx+2fy+c=0$ and $x^2+y^2+2g_{1}x+2f_{1}y+c_1=0$

Answer 1

Let the two circle be $x^2+y^2+2gx+2fy+c=0$

and $x^2+y^2+2g_{1}x+2f_{1}y+c_1=0$

If the circles intersect at $P$ then angle $\theta$ is

the angle between the tangents to both the circles at the point $P$.

$C_1$ and $C_2$ are centres of the circles.

$\Rightarrow C_1(-g,-f),C_2(-g_1,-f_1)$ and

Radius are given by $r_1=\sqrt{g^2+f^2-c},r_2=\sqrt{{g_1}^2+{f_1}^2-c_1}$

$d=\left|C_1{C}_2\right|=$distance between the centres

$\Rightarrow \sqrt{g^2+f^2+{g_1}^2+{f_1}^2-2g{g}_1-2f{f}_1}$

In $△C_{1}P{C}_2$

$\cos \alpha =\frac{r_1^2+r_2^2-d^2}{2r_1{r}_2}$

where $\alpha$ is the $\angle C_{1}P{C}_2$

$\theta$ is the angle $A^{\prime }P{B}^{\prime }$

$\alpha +\theta +{90}^{\circ }+{90}^{\circ }={360}^{\circ }$

$\Rightarrow \alpha ={180}^{\circ }-\theta$

So, $\cos ({180}^{\circ }-\alpha )=\frac{r_1^2+r_2^2-d^2}{2r_1{r}_2}$