Condition for intersection of chords inside a circle? What is the condition for intersection of 2 chords inside a circle?
Given n number of chords how to find the number of pairs of interecting chords?
 A: My try for $n=2$
Assume that the circle have ratio $r$ and is centered at $(0,0)$
Two chords with equations:
$$(r\cos{\theta_0},r\sin{\theta_0})+t(r\cos{\theta_1}-r\cos{\theta_0},r\sin{\theta_1}-r\sin{\theta_0}),0\leq t\leq 1$$
$$(r\cos{\theta_3},r\sin{\theta_3})+s(r\cos{\theta_3}-r\cos{\theta_2},r\sin{\theta_1}-r\sin{\theta_1}),0\leq s\leq 1$$
Have a intersection if:
$$(r\cos{\theta_0},r\sin{\theta_0})+t(r\cos{\theta_1}-r\cos{\theta_0},r\sin{\theta_1}-r\sin{\theta_0})=(r\cos{\theta_2},r\sin{\theta_2})+s(r\cos{\theta_3}-r\cos{\theta_2},r\sin{\theta_3}-r\sin{\theta_2})$$
Which is:
$$t(-\cos{\theta_0}+\cos{\theta_1})+s(\cos{\theta_2}-\cos{\theta_3})=\cos{\theta_2}-\cos{\theta_0}$$
$$t(-\sin{\theta_0}+\sin{\theta_1})+s(\sin{\theta_2}-\sin{\theta_3})=\sin{\theta_2}-\sin{\theta_0}$$
$$\begin{bmatrix}
     (-\cos{\theta_0}+\cos{\theta_1}) & (\cos{\theta_2}-\cos{\theta_3}) \\
     (-\sin{\theta_0}+\sin{\theta_1}) & (\sin{\theta_2}-\sin{\theta_3}) \\
  \end{bmatrix}\begin{bmatrix}
     t \\
     s \\
  \end{bmatrix}=\begin{bmatrix}
     \cos{\theta_2}-\cos{\theta_0} \\
     \sin{\theta_2}-\sin{\theta_0} \\
  \end{bmatrix}$$
The determinant is: 
$$2 \sin{\left(\frac{\theta_0-\theta_1}{2}\right)} \left(\cos{\left(\frac{ \theta_0+\theta_1-2 \theta_2}{2}\right)}-\cos{\left(\frac{ \theta_0+\theta_1-2 \theta_3}{2}\right)}\right)$$
which must be non zero and the solution must be between $0$ and $1$ for each component.
Note: if the determinant is zero we can note that one possibility is $\theta_3=\theta_2$ or another $\theta_1=\theta_0$ (one of the chords is a point)
