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There are plenty of ways you can represent a plane mathematically. One of the most succinct ones is using the normal and a position vector as $$\hat{n}\cdot (\vec{x}-\vec{c})=0$$ Where $\hat{n}$ is the vector normal to the plane and $\vec{c}$ is any vector in the plane. Now in your case you have two points and you know that the plane is vertical, which ...

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I am not sure I answer exactly the question as posted; so, forgive me if what I write is out of topic. If you have two curves defined by $y_1(x)$ and $y_2(x)$, the square of the distance between the two curves is given $$\Phi(x_1,x_2)=\big(x_1-x_2\big)^2+\big(y_1(x_1)-y_2(x_2)\big)^2$$ and you want to minimize this function with respect to $x_1$ and $x_2$. ...

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A common class of maps that takes lines to circular arcs are Möbius transformations. They are most commonly considered on complex numbers, but this translates to $\mathbb{R}^2$. A Möbius transformation of the (extended) complex plane is a map $z \mapsto \frac{a z + b}{c z +d}$ with $ad - bc \neq 0$.

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Imagine the smaller circle has rolled along the bigger circle as shown above. The distance it has rolled along the larger circle is equal to $bt$ (shown as the green path). The distance that point $m$ on the smaller circle has moved is equal to $as$ (shown as the pink path). These two distances must be equal, therefore:$$as=bt$$$$\therefore ... 1 The derivative of \nu is$$ \nu'(t) = (r'(t)\cos(t)-r(t)\sin(t),r'(t)\sin(t)+r(t)\cos(t)) $$so$$ \|\nu'(t)\|^2 =\dots=r'(t)^2+r(t)^2.  Therefore: $\nu'(t)=0\iff r(t)=r'(t)=0$. If $r'(t)\neq r(t)$ for all $t$, then $\nu'(t)\neq0$ for all $t$. (This condition is sufficient but not necessary.) The curve $\nu$ has speed one if and only if $r$ satisfies ...

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