(Following from: Difference between vectors with same origin?)
Given the parametric equations:
$x(t)=t+(X−t)cosθ$
$y(t)=(X−t)sinθ$
with $θ[0,2π)$ and $t>0$
where $X$ is a fixed point on the x axis.
The points with angle $θ$ can be seen as lying on a sphere centred at t with radius: $r=(X-t)$.
Below is a plot showing some of these points divided by angle, and the sphere they lie on.
For a fixed $X$, and a point (x,y) that satisfies the condition:
$x(t, θ)=t+(X−t)cosθ$
$y(t, θ)=(X−t)sinθ$
1) Prove that there is only 1 solution $(t, θ)$
for 2 points (x1,y1), (x2,y2) with fixed $X,θ$ such that:
$x1(t, θ)=t+(X−t)cosθ$
$y1(t, θ)=(X−t)sinθ$
$x2(t+1, θ)=(t+1=+(X−(t+1))cosθ$
$y2(t+1, θ)=(X−(t+1))sinθ$
To show that for each point $(x,y), x\ne X$, there is only one pair $t, \theta$ with $t > 0, 0 \le \theta <2\pi$ with $$x = t + (X - t) \cos \theta\\y = (X - t)\sin\theta$$ note that $$(x - t)^2 + y^2 = (X - t)^2\\t^2 - 2xt + x^2 + y^2 = t^2 - 2Xt + X^2\\x^2 + y^2-X^2 = 2(x - X)t\\t = \frac{x^2 + y^2-X^2}{2(x - X)}$$
(Excuse my comment above where I said there was also a solution with $t < 0$. I hadn't worked it all out in my head yet, and thought that the equation was quadratic in $t$.) For $\theta$, note that $\tan \theta = \frac{y}{x - t}$. There is only one such $\theta \in [0, 2\pi)$ that gives the correct signs for both $x - t$ and $y$.
For (2), I assume what you mean is $$\begin{align}(x_1,y_1) &= (t + (X - t) \cos \theta, (X - t)\sin\theta)\\ (x_2,y_2) &= (t + 1 + (X - t - 1) \cos \theta, (X - t - 1)\sin\theta)\end{align}$$ (Pro-tip: If you want to see how someone else produced a bit of mathematics, right-click on it and select "Show Math As > TeX Commands". Also, consult the MathJax Quick Reference.)
As I said in the comment, the quickest way to find the angle between any two real vectors (in any dimension) is by means of $$\vec v_1 \cdot \vec v_2 = \|v_1\|\|v_2\|\cos \Theta$$In this case, it will be helpful to let $q = (X - t)$ $$\begin{align}\vec v_1 \cdot \vec v_2 &= x_1x_2 + y_1y_2\\&= (t + q\cos \theta)(t + 1 + (q - 1)\cos \theta) + (q\sin \theta)((q - 1)\sin\theta)\\&= t(t+1) + (t(q - 1) + (t + 1)q)\cos\theta + q(q - 1)(\cos^2 \theta + \sin^2\theta)\\&= t(t+1) + (2qt + q - t)\cos\theta + q(q - 1)\end{align}$$ While $\|v_1\| = \sqrt{t^2 + q^2 + 2qt\cos\theta}, \|v_2\| = \sqrt{(t+1)^2 + (q-1)^2 + 2(q-1)(t+1)\cos\theta}$
So, $$\cos \Theta = \frac {t(t+1) + (2(X - t)t + X - 2t)\cos\theta + (X-t)(X - t - 1)}{\sqrt{\left(t^2 + (X - t)^2 + 2(X - t)t\cos\theta\right)\left((t+1)^2 + (X - t-1)^2 + 2(X - t -1)(t+1)\cos\theta\right)}}$$