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Let $p_0 = (x_0, y_0)$ be the center of the sector so that our circular sector is given by $S(p_0, r, \theta_1, \theta_2) = \{p_0 + r(\cos(\theta), \sin(\theta))\mid 0 \leq \theta_1\leq\theta\leq\theta_2 \leq 2\pi \}$. Furthermore let $p_1 \in \mathbb{R}^2\setminus \{p_0\}$ be a "destination" point, so that $p_0$ and $p_1$ are connected by a finite segment $l$. If our sector $S(p_0, r, \theta_1, \theta_2)$ moves along the segment $l(t) = p_0 + t(p_1 - p_0)$, what will be the resulting bounding curve for the shape? By "moving along" I mean that the end shape will be given by $\bigcup_{0 \leq t \leq 1}S(l(t), r, \theta_1, \theta_2)$.

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  • $\begingroup$ You may as well assume $p_0=(0,0)$ and $r=1$, without loss of generality. $\endgroup$
    – user519413
    Commented Nov 18, 2021 at 6:37

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I suppose that the vector $\vec{p_0p_1}$ does not lie in the cone spanned by the vectors defining the circular sector.

Let $q_0$ be the point defined by $p_0+r(\cos(\alpha), \sin(\alpha))$ with $\alpha = (\theta_1+\theta_2)/2$, i.e. the line $(p_0q_0)$ is a symmetry axis for the circular sector. The resulting shape of the sector ``moved along'' the segment $[p_0p_1]$ is obtained by replacing the segment $(p_0q_0)$ with the parallelogram with vertices $p_0$, $q_0$, $q_1$, and $q_0$, where $q_1 = p_1 +(q_0 -p_0)$. The bounding curve has the same geometric description. In case a parametrization is needed, setting $\gamma(t) = (\cos\theta, \sin\theta)$, we might take $$p(t) = \begin{cases} p_0 +tr\gamma(\theta_1) & t \in [0, 1] \\ p_0 +r\gamma((t-2)\theta_1 +(t-1)\alpha) & t \in [1, 2] \\ q_0 +(t-2)(p_1 -p_0) & t \in [2, 3] \\ p_1 +r\gamma((t-4)\alpha +(t-3)\theta_2) & t \in [3, 4] \\ p_1 +(5-t)r\gamma(\theta_2) & t \in [4, 5] \\ p_1 +(t-5)(p_0 -p_1) & t \in [5, 6]. \end{cases}$$

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