I have a circle with radius $r$ and a line coming from the center of that circle at a distance $L_0$ like this: enter image description here

then, I want to find the distance $L$ to a different point on this circle, measured by angle $\theta$ which goes clockwise, like this:

enter image description here

Is there any way I can calculate $L$ as a function of $\theta$ and $L$? I tried doing some trigonometry but was unsuccessful.

What about in the more general case where the initial point of $L_0$ is not perpendicular to the circle? For example, this:

enter image description here


You could use the Law of Cosines (which you might think of as a super-powered version of the Pythagorean Theorem, which allows you to work with triangles that are not right triangles). It states that in any triangle with angles $a,b,c$ opposite sides of length $A,B,C$ (respectively), we have $$ A^2 = B^2 + C^2 - 2AB \cos(a). $$ In your problem, we know the three sides, and so we get $$ L^2 = r^2 + (r+L_0)^2 - 2r(r+L_0) \cos(\theta) \implies \cos(\theta) = \frac{L^2 - r^2 - (r+L_0)^2}{2r(r+L_0)}.$$ Therefore, up to a choice of quadrant, $$ \theta = \arccos\left( \frac{L^2 - r^2 - (r+L_0)^2}{2r(r+L_0)} \right). $$

  • $\begingroup$ That all makes sense, I actually was thinking of a more general case, which I added a diagram of in my question. I should probably submit a new question with just the more general diagram. $\endgroup$ – TheStrangeQuark Aug 2 '17 at 14:39
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    $\begingroup$ Look for triangles that will help you get the job done, and remember the various congruence relations. If you know $\varphi$, $L$, and $L_0$, then the "side-angle-side" congruence relation indicates that you can find the third side of the triangle formed by $L$ and $L_0$ (e.g. using the Law of Sines). If we call this side $A$, you can then consider the triangle with sides $A$, $r$, and $r$ and the angle $\theta$. Using the Law of Cosines, you can find $\theta$. $\endgroup$ – Xander Henderson Aug 2 '17 at 14:46
  • $\begingroup$ But I don't know $L$ or $\phi$. All I know is $L_0$ and $\theta$. Is it even possible to calculate $L$ with this little information? $\endgroup$ – TheStrangeQuark Aug 2 '17 at 16:44
  • $\begingroup$ The segment $L$ is not uniquely determined by only $\theta$ and $L_0$. If you fix one end of $L_0$ on the circle, but allow the other end to vary, it will trace out another circle. Each point on that circle can be joined to a point on the original circle, giving a different possible segment $L$. $\endgroup$ – Xander Henderson Aug 2 '17 at 17:07


$$L^2=(L_0+r)^2+r^2-2r (L_0+r)\cos (\theta) $$

$$=L_0^2+4r(r+L_0)\sin^2 (\frac {\theta}{2})$$

  • $\begingroup$ To be fair, that's not really a hint. That's the answer. $\endgroup$ – T. Linnell Aug 2 '17 at 14:26

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