Parametric equation of an arc with given radius and two points so I need the parametric equation of the arc. So, arc is a sector of a circle.
Parametric circle equation is:
$$
c \equiv f(t) = (\cos(t), \sin(t)),\quad 0\le t < 2\pi
$$
So, we just need to find proper domain of the function, actually $t_1$ and $t_2$, start and end of a sector.
Given two points $P_1$ and $P_2$, liying on circle, its center and radius how to find $t_1$ and $t_2$ using given points?
I need full parametric equation of this.
Thanks in advance!

Late answer, but I think I can add something useful to the discussion. When radius is given, you have to choose between 2 possible centers and then between 2 possible arcs. Instead, given the oriented angle $\theta$ (counterclockwise if positive or clockwise if negative), there is only one possible arc going from point $P_1=[x_1,y_1]$ to point $P_2=[x_2,y_2]$. In such case you don't have to choose anything, and you can obtain center $C$ and radius $r$ from the following expression (I used Maxima to obtain them):
$$\mathit{c_x}=\frac{\left(1+ \cos{(\theta)}\right) \, \left( \mathit{y_1}-\mathit{y_2}\right) +\sin{(\theta)} \left( \mathit{x_1}+\mathit{x_2}\right) }{2 \sin{(\theta)}}$$
$${c_y}=\frac{\sin{\left( \theta \right) \left( {y_1}+{y_2}\right)}+\left( 1+\cos{\left( \theta \right) }\right)\left( {x_2}-{x_1}\right)   }{2 \sin{\left( \theta \right) }}$$
$$r=\sqrt\frac{{{\left( \mathit{x_2}-\mathit{x_1}\right) }^{2}}+{{\left( \mathit{y_2}-\mathit{y_1}\right) }^{2}}}{2 \left( 1-\cos{(\theta)}\right) }$$
