How do I calculate distance between origin and point on a chord in the unit circle I have a chord connecting two points $P_1$ and $P_2$ on the unit circle. I would like to calculate the distance $(l_1,\,l_2,\, l_3, \ldots)$ between the origin $(0,0)$ and the point on the chord for "every" point on the chord. I have the arc-length and I set/choose angle theta. 
I don't have the $x$ and $y$ coordinates of the singular points on the chord only the coordinates of $P_1$ and $P_2$.
A picture for clarification:

Thank you
 A: I see two ways to consider the issue (A and B) :
A) If the coordinates of points $P_k$ are $(\cos \theta_k, \sin \theta_k)$ for $k=1,2$, consider the barycentric expression :
$$\binom{x}{y}=(1-\lambda) \binom{\cos \theta_1}{\sin \theta_1}+\lambda\binom{\cos \theta_2}{\sin \theta_2} \tag{1}$$
taking all values of $\lambda$ between $0$ and $1$, you will browse all your chord.
Said otherwise, each point of the chord will be characterized by a unique $\lambda$.
Therefore the "generic" distance to the origin will be :
$$d_{\lambda}=\sqrt{x^2+y^2}$$
with $(x,y)$ given by (1).
Remark : In order to understand formula (1), two  examples,

*

*taking $\lambda = \frac12$ you get the midpoint of your chord,


*taking $\lambda = \frac34$, you are 3 times closer to $P_2$ than to $P_1$
Besides, if you eliminate $\lambda$ between the two expressions $x=..., y=...$,  you will get the equation of the straight line, but I don't think that it is what you want.

B) Here is a different way.
Call $I$ the midpoint of your chord.
Let $(r_I,\theta_I)$ be the polar coordinates of $I$ (meaning in particular that distance $OI=r_I$.
If we consider now polar coordinates $(r_M,\theta_M)$ of any point $M$ on the chord, we have the simple relationship:
$$dist(O,M)=r_M=\dfrac{r_I}{\cos(\theta_I-\theta_M)}\tag{2}$$
(proof : definition of $\cos$ in right triangle $OIM$).
A: You know that for any point in the plane, $$\tan\theta=\frac yx$$
The equation of the line is $$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}$$
Here $P_i=(x_i,y_i);\ i=1,2$.You have a sytem of two linear equations, with two variables. Find $x$ and $y$, square them, add the squares and you get $l^2(\theta)$.
