finding center of circle How can I calculate center of a circle $x,y$? I have 2 points on the circumference of the circle and the angle between them.
The 2 points on the circle are $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$. The angle between them is $\theta$. I know all these values; how can I calculate the center from these values?
 A: 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)...
A: It's quite easy using complex numbers. 
You can represent every point $(x,y)$ through the complex number $x+iy$ and vice versa. It's just easier to calculate with them this way. 
Let $P_1 = 0$, then we get $S = \frac {P_2} 2$ as the starting point on the bisector. Multiply by $i$ to rotate by 90°, thus we need to go in the direction of $i\frac{S}{|S|}$. The distance is $d = \frac{|S|}{\tan \frac \theta 2}$, thus we get the center by
$$M = S + i\frac{S}{|S|}\cdot d = S(1 + \frac i {\tan \frac \theta 2})$$
Now for arbitrary points $P_1, P_2$, we get
$$ M = \frac 1 2 (P_2-P_1) (1 + \frac i {\tan \frac \theta 2}) + P_1. $$
A: Hint: 
Find the perpendicular bisector of $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$.
A: Angle at centre is twice the angle between the given points
let centre be (x,y)  
form equations for line segments joining centre to these 2 points respectively
you also have the ansgle between them, so solve to get (x,y)
A: The bisector of the line segment $\overline{P_1P_2}$ passes through the circle's center.
Let $\alpha = \frac \theta 2$ and you get the distance $d$ to travel on the bisector by $$\tan \alpha = \frac{\overline{P_1P_2}}{d}.$$
