# equation to get 10 points on circle surface at fix distance

What I tried is :

$$x = \sin(36 \cdot 50 \cdot 3.14)/180$$

$$y = \cos(36 \cdot 50 \cdot 3.14)/180$$

Here

$36$ is because I want 10 points on circle so $360/10=36$.

$50$ is center X and center Y of circle. That is center point of circle is $(50,50)$.

$3.14$ is value of $\pi$

$180$ is again value of $\pi$ in radian

But it always returns $0,0$. How can I get $10$ points on circle surface at fix distance

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If you can specify your points in polar coordinates, you're golden… – Seamus Feb 29 '12 at 15:31
@mark Find all tenth roots of unity and plot them in the complex plane. – Pedro Tamaroff Mar 6 '12 at 3:21

Your formulas are off.

You want the angles $$36^\circ,\ 2\cdot36^\circ,\ 3\cdot36^\circ,\ \ldots,\ 10\cdot36^\circ.$$

To convert degrees to radians, multiply by ${\pi\over 180}$.

So the angles in radians are $${36\pi\over 180} ,\ {2\cdot36\pi\over 180} ,\ {3\cdot36\pi\over 180} ,\ \ldots, {10\cdot36\pi\over 180} .$$

With center $(50,50)$ and radius $r$, the point $(x,y)$ on the circle is $$x=50+r \cos\theta, \quad y=50+r \sin\theta.$$ where $r$ is the radius of the circle (you didn't specify $r$).

So for the 36 degree angle, for example, the point is $$\textstyle x=50+r\cos( {36\pi\over 180} ),\quad y=50+r\sin( {36\pi\over 180}\cdot36 ).$$ or, simplifying $$\textstyle x=50+r\cos( { \pi\over5} ),\quad y=50+r\sin( { \pi\over5} ).$$

(Note the angles are just ${\pi\over 5}$, ${2\pi\over 5}$, $\ldots\,$, ${10\pi\over 5}$.)

You could save some computational effort by taking advantage of symmetry:

The three points in the "first quadrant" of the circle are (corresponding to the angles ${10\pi\over5}=2\pi\sim0$, ${\pi\over 5}$, and ${2\pi\over5}$):

$$\textstyle ( 50+r,50 ),\quad (50+r\cos{\pi\over5}, \quad 50+\sin{\pi\over5}),\quad (50+r\cos{2\pi\over5}, 50+\sin{2\pi\over5})$$

The three points in the "second quadrant" of the circle are (corresponding to the angles ${5\pi/5}=\pi$, ${3\pi\over 5}$, and ${4\pi\over5}$):

$$\textstyle ( 50-r,50 ),\quad (50-r\cos{\pi\over5}, \quad 50+\sin{\pi\over5}),\quad (50-r\cos{2\pi\over5}, 50+\sin{2\pi\over5})$$

And to get the other four points in the "bottom half" of the circle, take the points above, except $(50+r,50)$ and $50-r,50)$, and switch the sign before the "$\sin$" in the $y$ coordinates.

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