Overview: Given an arbitrary point $t$ on the horizontal axis of a Cartesian plane and a point $\textbf{p}$ on a circle, I would like to find the point $\textbf{t}'$ located at the intersection of the line passing through $(t,0)$ and $\textbf{p}$, and the circle (assuming Euclidean space).
Edit: in light of the answer given by Aretino, my derivation here is extremely clunky. I have left it in for context, but it is otherwise superfluous.
Let $\textbf{f}:\mathbb{R}\to\mathbb{R}^2$ s.t. for a circle $C$ with center $\textbf{c}$ and pole $\textbf{p}$, $\textbf{f}(t)$ is the intersection of a line segment from $(t,0)$ to $\textbf{p}$ and $C$ - i.e. $t$ is a [stereographic] projection of $\textbf{f}(t)$, and $\textbf{f}(t)$ is a parameterization of $C$.
$C$ is described implicitly by the equation
$$(x_1-c_1)^2+(x_2-c_2)^2=r^2$$
where $r$ is the radius of $C$
If the axis of $C$ (the line passing through both $\textbf{c}$ and $\textbf{p}$) is perpendicular to the horizontal ($x_2=0$), then:
$$\textbf{f}(t)=\left(\frac{2r(t-c_1)}{\frac{(t-c_1)^2}{p_2}+p_2}+c_1\ ,\ r\frac{\frac{(t-c_1)^2}{p_2}-p_2}{\frac{(t-c_1)^2}{p_2}+p_2}+c_2\right)$$
What is the equation for $\textbf{f}$ given an arbitrary axis?
Note: if the axis and the horizontal coincide, then $\textbf{f}(t)=\begin{cases}(c_1-r,0)&t<c_1\\(c_1+r,0)&t>c_1\end{cases}$. The projection goes from being a 1-sphere to a 0-sphere.
We have:
$$ |\mathbf{p}-\mathbf{t}|\cdot|\mathbf{p}-\mathbf{t'}|=|\mathbf{c}-\mathbf{t}|^2-r^2, $$ hence: $$ \mathbf{t'}=\mathbf{t}+{\mathbf{p}-\mathbf{t}\over|\mathbf{p}-\mathbf{t}|} |\mathbf{p}-\mathbf{t'}|= \mathbf{t}+(\mathbf{p}-\mathbf{t}){|\mathbf{c}-\mathbf{t}|^2-r^2\over|\mathbf{p}-\mathbf{t}|^2}. $$ Where $\mathbf{t}$ is located on the horizontal axis, $\mathbf{t}=(t,0)$. Thus, in coordinates: $$ t'_1=t+(p_1-t){(c_1-t)^2+c_2^2-r^2\over(p_1-t)^2+p_2^2},\\ t'_2=p_2{(c_1-t)^2+c_2^2-r^2\over(p_1-t)^2+p_2^2}.\\ $$