Set the parametric equation of an arc with two points Like the title says, I am looking for a method to find the parametric equation expressed in the form  $\vec{r}=...\vec{i}+...\vec{j}$ of the arc that connects the points (2,0) and (1,2). I am asking for this, because it is the step that is stopping me to compute the value of a vector field.
Thank you.
 A: Try $r(t)=(-t,2t)+(2,0)=(2-t,2t)=(2-t)\ \vec i+2t\ \vec j$ with $0\le t\le 1$.
A: Several commenters have noted that the arc you want to define is not unique; there are an infinity of such arcs. More precisely, there is a natural parametrization of such arcs by a single variable and a choice of orientation, which I describe below.
When you draw the situation on paper, you will see that the center of the circle must exist on the line which is the perpendicular bisector of the line segment between the two points. We can prove this: if the two given points are $A$ and $B$, their midpoint is $C$, and $O$ is the center of such a circle, then:


*

*$OA$ and $OB$ must have the same length

*$\Delta OAB$ is isosceles

*$\Delta OCA\cong \Delta OCB$ by side-angle-side [this one you probably have to draw to see]

*$\angle OCA \cong \angle OCB$

*Since $A,B,C$ are colinear, the measures of those angles must sum to $180^\circ$.

*Since they are equal, $m\angle OCB = \left(\frac{180}2\right)^\circ$

*Thus, $O$ lies on the line perpendicular to $CB$ and passing through $C$, as desired.


Now we can find an explicit representation of this line; I will use vector notation. The line between $A=(a_0,b_0)$ and $B=(a,b)$ is given by this one-variable ($t$) vector equation:
$$ \begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix} a-a_0\\b-b_0 \end{bmatrix} t + \begin{bmatrix} a_0\\b_0 \end{bmatrix}. $$
Since $C=(\frac{a+a_0}{2},\frac{b+b_0}{2})$, we have that the perpendicular line is given by
$$ \begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix} b_0-b\\a-a_0 \end{bmatrix} t + \begin{bmatrix} \frac12(a+a_0)\\\frac12(b+b_0) \end{bmatrix}. $$
So each value of $t$ gives us a possible center for our circle.
At this point, we recall that a circle is given by $(x-h)^2+(y-k)^2=r^2$, where the center $O=(h,k)$ and the radius ($OA$, or $OB$) has length $r$. For a particular choice of $t$:
\begin{align*}
h &=(b_0-b)t+(a+a_0)/2 &&= -\beta t + \alpha/2 \\
k &=(a-a_0)t+(b+b_0)/2 &&= \alpha t + \beta/2 \\
r^2 &= \displaystyle \bigg[ (b_0-b)^2t^2 + (a-a_0)^2t^2\bigg] + \bigg[ \frac{(a-a_0)^2}{4} + \frac{(b-b_0)^2}{4} \bigg] &&= \textstyle(\alpha^2+\beta^2)(t^2+\frac14)
\end{align*}
where $\alpha=a-a_0$ and $\beta=b-b_0$. The calculation of $r^2$ comes from using the Pythagorean theorem on the triangle $\Delta OCA$: the first square-bracket is the length of $OC$ and the second is the length of $CA$ (using the distance formula).
We now pull all the data together to get a circle: You have $(a_0,b_0)=(2,0)$ and $(a,b)=(1,2)$ which means $\alpha=-1$ and $\beta=2$; thus
$$\textstyle(x-2t-\frac12)^2 + (y+t+1)^2 = \frac54 + 5t^2.$$
Now for the usual circle trick: Let $\theta$ be defined such that
$$\cos\theta = \frac{x-2t-\frac12}{\sqrt{\frac54+5t^2}} \qquad\qquad\text{and}\qquad\qquad \sin\theta = \frac{y+t+1}{\sqrt{\frac54+5t^2}} ~.$$
This means that this circle has parametric form
$$ \begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix} 2t+\frac12\\-t-1 \end{bmatrix} + \sqrt{\frac54+5t^2}\begin{bmatrix} \cos\theta\\\sin\theta \end{bmatrix}. $$
[Remeber: $\theta$ is the only parameter here; the $t$ is just showing up because the circle is not uniquely defined. For each available arc, $t$ is just a specific constant number.]
Therefore, the only remaining task is to determine which range of $\theta$ will create the arc with endpoints $A$ and $B$. This can be done in general though a case-by-case analysis of possible $t$-values, but it is very tedious. The reason that it is not so simple to do it directly is that plugging in $x=2$ to the $\cos\theta=\cdots$ equation does not uniquely define $\theta$, even on the interval $[0,2\pi)$; one must also know at least the sign of $\sin\theta$ when $y=0$, and hence there are two cases to consider. A similar analysis happens with the parameters for $B$ inserted, so that there are now four cases to consider. 
Finally, you have some flexibility to choose the $\theta$ values you choose for the endpoints, and the choices you make at this step are not totally immaterial. Of course you have freedom to choose either one up to adding a multiple of $2\pi$: the point is that by choosing the same multiple or consecutive multiples, and depending on which of the four cases you are in, you will change whether you get a major or minor arc.
