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


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: $$ t'_1=t+(p_1-t){(c_1-t)^2+c_2^2-r^2\over(p_1-t)^2+p_2^2},\\ t'_2=p_2{(c_1-t)^2+c_2^2-r^2\over(p_1-t)^2+p_2^2}.\\ $$

  • $\begingroup$ I'm not sure I follow. Is $\textbf{t}=(t,0)$? If so, then wouldn't $t'_2$ end up being constant? $\endgroup$ – R. Burton Mar 11 at 23:13
  • $\begingroup$ It isn't constant. I can write the formula with coordinates if it's not clear. $\endgroup$ – Aretino Mar 11 at 23:21
  • $\begingroup$ That might help. Is this from the line to a circle or from a plane to a sphere? $\endgroup$ – R. Burton Mar 11 at 23:25
  • $\begingroup$ It's all in the plane. $\endgroup$ – Aretino Mar 11 at 23:28
  • $\begingroup$ In fact I'm assuming the circle lies all on the same side of the horizontal axis, that's where my reasoning may fail. $\endgroup$ – Aretino Mar 11 at 23:39

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