Finding angle between two points on circle given cartesian coordinates of the points If I am given the coordinates of two points on a circle of radius'R', how can I find the angle between those two points, as well as the area of the arc between them(created on the circumference of the circle).  
 A: If two points $P_0(x_0,y_0)$ and $P_1(x_1,y_1)$ on the circle and the radius of the circle $R$ are given, then calculate the coordinates of the middle point
$$P_M=(x_M,y_M)=((x_0+x_1)/2,(y_0+y_1)/2)$$
between $P_0$ and $P_1$.
The radius $R$ of the circle is perpendicular to the line that goes through the points $P_0$ and $P_1$, if we draw it through the point $P_M$. Then calculate the distance between $P_0$ and $P_M$
$$d(P_0,P_M)=\sqrt{(x_0-x_M)^2+(y_0-y_M)^2}=(1/2)\cdot\sqrt{(x_0-x_1)^2+(y_0-y_1)^2}$$
or the distance between $P_1$ and $P_M$
$$d(P_1,P_M) =(1/2)\cdot\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}$$
by the pythagorian theorem and note that these distances are equal,  $d(P_0,P_M)=d(P_1,P_M)$. Now the angle $\phi$ can be calculated from
$$\tan(\phi /2)=d(P_0,P_M)/(R-\color{green}{f})=d(P_1,P_M)/(R-\color{green}{f})$$

You can use the formula described here to calculate $\color{green}{f}$.
EDIT
or you can use the pythagorian theorem again:
Let
$$d = d(P_0,P_M) = d(P_1,P_M),$$ then $$\tan(\phi /2)=d/\sqrt{R^2-d^2}$$
and finally
$$\phi = 2\cdot \arctan(d/\sqrt{R^2-d^2})$$
Putting it all together, we get
$$\phi = 2\cdot \arctan\left(     \frac{\sqrt{(x_1-x_0)^2+(y_1-y_0)^2} }{\sqrt{4R^2- (x_1-x_0)^2-(y_1-y_0)^2 }}\right)$$
$$\phi = 2\cdot \arccos\left(     \frac{\sqrt{4R^2-(x_1-x_0)^2-(y_1-y_0)^2} }{2R }\right)$$
$$\phi = 2\cdot \arcsin\left(     \frac{\sqrt{(x_1-x_0)^2+(y_1-y_0)^2} }{2R }\right)$$
A: If we take the dot product of the vectors from the center to these points then, $u\cdot v = |u||v|\cos \theta$  where theta is the angle you seek.
And since this is a circle $|u| = |v| = R$
If your circle is centered at the origin, and your points are $(x_1,y_1),(x_2,y_2)$
$\theta = \arccos \frac {x_1x_2 + y_1y_2}{R^2}$
