Given a circle $C$ with the equation $(x-a)^2 + (y-b)^2 = r^2$ and two points $(x_0,y_0), (x_1,y_1)$ on the circle, how can we find the continuous path $p:[0,1] \to \mathbb{R}^2$ from $(x_0,y_0)$ to $(x_1,y_1)$ which its graph is on the circle?
i.e $p([0,1])\subseteq C$, $p(0) = (x_0,y_0), p(1) = (x_1,y_1)$
I really have no clue where to start, I only know it has something to do with topology.
This is for animation purposes.
I shall outline a construction of the required map $p: [0,1] \to C \subset \Bbb R^2$. The given circle $C$ has center $(a,b)$ and radius $r$, and so any point on $C$ can be parametrized as $(a + r\cos\phi, b + r\sin \phi)$ where $\phi \in [0, 2\pi)$. Suppose $(x_0,y_0) = (a + r\cos\theta_0, b + r\sin \theta_0)$ and $(x_1,y_1) = (a + r\cos\theta_1, b + r\sin \theta_1)$. This gives $$\tan\theta_i = \frac{y_i - b}{x_i - a}$$ for $i = 0,1$, enabling one to calculate $\theta_0$ and $\theta_1$. Consider the map $p: [0,1] \to \Bbb R^2$ given by $$p(t) = (a + r\cos (\theta_0 + t(\theta_1 - \theta_0)), b + r\sin (\theta_0 + t(\theta_1 - \theta_0))).$$ Clearly, the image of $p$ lies in $C$, $p$ is continuous, $p(0) = (x_0,y_0)$ and $p(1) = (x_1,y_1)$.