Condition for intersection of chords inside a circle?

Question

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?

Answer 1

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)