# Finding the center of an n-gon (circle) based on angle and side-length

by STENDEC
Tags: angle, based, circle, ngon, sidelength
 P: 14 I hope this is self-evident to someone, i'm struggling. I have a program that draws circles (n-gons really) of various sizes, but by translating-rotating-translating-rotating-..., not by x=sin/y=cos. That works as intended, but my wish is to offset the circle so that its center is (0,0) in the coordinate system. For that i need its center. Currently the circle itself originates from- and hence touches the (0,0) coordinates, so its center is somewhere above, in the y-axis. Position of ? is sought after. A wider angle would result in ? rising for instance. I found lots of tutorials on how to do it on paper using dividers and i also considered that it's a isosceles triangle, but it seems all textbook examples assume that one of the symmetric sides is already known.
 P: 2,991 you could use similar triangles and some trig to get the radius along the y-axis. Notice you can extend a perpendicular bisector from the first n-gon side which intersects the y-axis so that 1/2 the n-gon side is the short edge the perpendicular creates the right angle and the y-axis is the hypotenuse. This triangle is similar to the one formed by the n-gon edge and the x-axis. So I get something like: radius along y-axis = (1/2 n-gon side) / sin theta
P: 14
 Quote by jedishrfu you could use similar triangles and some trig to get the radius along the y-axis.
Yes, you're right. After some more reading and pondering i came to this solution:

$\alpha =$ angle in degrees
$s =$ segment length

To get the inner angle between the sides, we subtract from a half-circle. We then divide by two, to get the inner angle of the isosceles triangle:
$\beta = (180 - \alpha) \div 2$

$\phi = \beta\times\frac\pi{180}$

Distance to center point can then be gotten from $s\div 2 * tan(\phi)$.

Edit: Just saw you extended your reply, oh well :)

P: 2,991
Finding the center of an n-gon (circle) based on angle and side-length

 Quote by STENDEC Yes, you're right. After some more reading and pondering i came to this solution: $\alpha =$ angle in degrees $s =$ segment length To get the inner angle between the sides, we subtract from a half-circle. We then divide by two, to get the inner angle of the isosceles triangle: $\beta = (180 - \alpha) \div 2$ degrees to radians: $\phi = \beta\times\frac\pi{180}$ Distance to center point can then be gotten from $s\div 2 * tan(\phi)$. Edit: Just saw you extended your reply, oh well :)
Glad you figured it out.

 Related Discussions Calculus 1 Differential Geometry 3 Precalculus Mathematics Homework 1 General Math 7 Introductory Physics Homework 10