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I have a circle that I am trying to find series of points on. I know the radius and horizontal tangent point at the top of the circle. I need to find a point that lies on the circle's circumference that is $x$ distance below the top point.

2 Answers2

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The vertical coordinate of that point will obviously be $Y=R-x$, where $R$ is the radius. The horizontal one follows from Pythagoras as $X=\sqrt{R^2-(R-x)^2} = \sqrt{2Rx - x^2}$. I don't see where you need this tangent for? Or is this by construction?

Lucozade
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  • I am trying to write a program that fills a circle with horizontal lines at an even distance. The lines start at the top of the circle and continue down. I know the point that is in the middle of the circle and at the very top. I need to calculate the points on the circumference every second x points down from that point. –  Jul 03 '13 at 18:58
  • Your equation solved my problem. It took a little bit for me to realize that I needed to use the negative of that equation to solved for the points on the right side of the triangle though. –  Jul 03 '13 at 22:45
  • I meant circle, not triangle. –  Jul 04 '13 at 14:54
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Let the equation of the circle be $$(x-h)^2+(y-k)^2=r^2$$

So, the horizontal tangent point at the top of the circle will be $(h,k+r)$

So, we know $h,k,r$

Any point $(P)$ on the circle can be $(h+r\cos\theta,k+r\sin\theta)$

So, $(h+r\cos\theta-0)^2+\{k+r\sin\theta-(k+r)\}=x^2$

So, we need to find $\theta$

lab bhattacharjee
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  • I thought of rcos(theta) and rsin(theta) immediately, but I cannot work with angles - see my comment to @Lucozade. –  Jul 03 '13 at 18:59