circlesgeometry
I have two points on the circumference of circle, and I also know the center of the circle. I want to calculate the angle between those two points which are on the circumference of circle.
Is this formula suitable to this situation?
$$\tan(\theta) = \frac{y_2-y_1}{x_2-x_1}$$
where $(x_1,y_1)$ are one of the circumference points and
$(x_2,y_2)$ is the other point on the circumference.
You could:
First, find the distance between, and the midpoint of, the two points.
Then draw the right triangle formed one of the given points, $p_1$, the midpoint, $m$, and the center, $c$, of the circle (this will be a right triangle since a perpendicular bisector of a chord on a circle passes through the center of the circle).
You know the length of the side $\overline{p_1m}$ of this triangle, since you know the distance between the given two points. You also know that the angle $\angle mcp_1$ is half the given angle.
Now, a bit of trig will allow you to find the side length $\overline{cm}$ of the triangle. Let's call that length $l$.
Next, find the equation of the line containing $\overline{cm}$ (it's slope will be the negative reciprocal of the slope of the line segment joining the two given points).
Let's say that equation is $y=m_0x+b$.
If $(c_1,c_2)$ are the coordinates of the center and $(m_1,m_2)$ are the coordinates of the midpoint, you'd know: $$ l = \sqrt{(c_1-m_1 )^2+(c_2-m_2)^2 } $$ and $$ c_2=m_0 c_1+b. $$ Finally, you'd solve the above equations for $c_1$ and $c_2$.
(There are probably slicker ways to do this.)
As Ross Millikan points out, there are two solutions (hence, there is another somewhat different diagram for the solution not represented by the above diagram)...
Let $d$ be the (straight-line) distance between the two points. Then the arclength between them is $$ s = 2r\sin^{-1}\left( \frac{d}{2r} \right) $$ Note that this does not assume that the circle is centered at the origin (as some of the other answers seem to do).
Of course, the problem only makes sense if $d \le 2r$, for otherwise there can be no circle passing through the two given points.
As mentioned in a comment, on any given circle passing through the two points, there is a shorter arc and a longer arc. The formula above gives the arclength $s$ of the shorter arc. The arclength of the longer one is simply $2\pi r - s$.
Best Answer
You have an isosceles triangle.
You can use cosine formula for calculation the angle.
$$c^2 = a^2 + b^2 -2ab \cos(\alpha)$$
$a$ and $b$ are sides next to the angle $\alpha$, which are the radius of the center $r$. $c$ is the distance between the two points $P_1$ and $P_2$. So we get:
$$\left|P_1 - P_2\right|^2 = 2r^2-2r^2 \cos(\alpha)$$
$$\frac{2r^2-\left|P_1 - P_2\right|^2}{2r^2} = \cos(\alpha)$$
$$\alpha = \cos^{-1}\left(\frac{2r^2-\left|P_1 - P_2\right|^2}{2r^2}\right)$$