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I know how to find a point on a circle given a radius and an angle, but my knowledge of trigonometry doesn't extend much further than that. My question is probably best explained diagrammatically:

enter image description here

Given a centre point, radius, and two points on the circle, how do I find the 'middle' point between those two points? I'm also interested in knowing the distance between the internal line formed between the two points and this new middle point. (Referring to the diagram, I'm trying to determine the distance, q, and the unknown point, x?, y?)



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The center point needs to be relabeled. – Matt Calhoun Mar 7 '11 at 15:39

First, you have two points $(x_1,y_1)$ in your diagram. I will use that for the one on the circle. You can calculate $d$ from the coordinates of $(x_1,y_1)$ and $(x_2,y_2)$ as $d=\frac{1}{2}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. If you draw the line segment from the center to $(x_1,y_1)$ you have a right triangle with sides $d, (r-q)$, and $r$, so $r^2=d^2+(r-q)^2$ which you can solve for $q$ by the quadratic equation. Then the coordinates of the intersection of the radius and the chord are $(\frac{x_2+x_1}{2},\frac{y_2+y_1}{2})$ and you can extend the line segment from the center through that until it has length $r$

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Write $(x_1,y_1)=(Rcos\theta_1,Rsin\theta_1)$ and $(x_2,y_2)=(Rcos\theta_2,Rsin\theta_2)$, where $R = \sqrt{x_1^2+y_1^2}$ is the radius of the circle, and solve for $\theta_1$ and $\theta_2$; then clearly $(x_?,y_?)=(Rcos(\theta_1+\delta),Rsin(\theta_1+\delta)$, where $\delta = \frac{\theta_1-\theta_2}{2}$.

Now that we have all the $(x,y)$ coordinates for all the relevant points, it's easy to set up the linear system of 2 equations with two unknown's, and to solve for the point of intersection of the two lines. The two equations are of course the equations for two lines drawn in the figure.

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