# Explicit formula for the projection from the line to an arbitrary circle

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 $$\textbf{f}(t)=\begin{cases}(c_1-r,0)&tc_1\end{cases}$$. The projection goes from being a 1-sphere to a 0-sphere.

$$|\mathbf{p}-\mathbf{t}|\cdot|\mathbf{p}-\mathbf{t'}|=|\mathbf{c}-\mathbf{t}|^2-r^2,$$ 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: $$t'_1=t+(p_1-t){(c_1-t)^2+c_2^2-r^2\over(p_1-t)^2+p_2^2},\\ t'_2=p_2{(c_1-t)^2+c_2^2-r^2\over(p_1-t)^2+p_2^2}.\\$$
• I'm not sure I follow. Is $\textbf{t}=(t,0)$? If so, then wouldn't $t'_2$ end up being constant? – R. Burton Mar 11 at 23:13