# Find the angle of intersection of two circles?

$\begin{array}{1 1}(a)\;\cos(180^{\large\circ}-\theta)=\large\frac{r_1^2+r_2^2-d^2}{2r_1r_2}\\(b)\;\cos(180^{\large\circ}+\theta)=\large\frac{r_1^2-r_2^2-d^2}{2r_1r_2}\\(c)\;\cos(180^{\large\circ}-\theta)=\large\frac{r_1^2-r_2^2+d^2}{r_1r_2}\\(d)\;\cos(180^{\large\circ}-\theta)=\large\frac{r_1^2+r_2^2+d^2}{4r_1r_2}\end{array}$

Toolbox:
• In a $\Delta\:ABC$ $cosA=\large\frac{b^2+c^2-a^2}{2bc}$
Let the two circles be $x^2+y^2+2gx+2fy+c=0$ and $x^2+y^2+2g_1x+2f_1y+c_1=0$
If the circles intersect at P then angle $\alpha$ 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 $r_1=\sqrt{g^2+f^2-c},r_2=\sqrt{g_1^2+f_1^2-c_1}$
$d=|C_1C_2|$=distance between the centres
$\Rightarrow \sqrt{g^2+f^2+g_1^2+f_1^2-2gg_1-2ff_1}$
In $\Delta C_1PC_2$
$\cos \alpha=\big(\large\frac{r_1^2+r_2^2-d^2}{2r_1r_2}\big)$
where $\alpha$ is the angle $C_1PC_2$
$\alpha+\theta+90^{\large\circ}+90^{\large\circ}=360^{\large\circ}$
$\theta=180^{\large\circ}-\alpha$
So $\cos(180^{\large\circ}-\theta)=\large\frac{r_1^2+r_2^2-d^2}{2r_1r_2}$
Hence (a) is the correct answer.
edited Mar 27, 2014