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I'd like to calculate the following angle, $\phi$ between the center of a circle $O$ and a point on the circumference $Q$ (wrt the $x$ axis), where $Q$ is determined by 2 parameters, $a$ and $\theta$, as defined in the diagram below. ($a$ is given as the fraction of the radius.)

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Would anyone be able to help determine an analytic expression for $\phi$ in terms of $a$ and $\theta$?

Would be extremely grateful for any help. Thank you.

Hosam Hajeer
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dogbiscuit123
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  • Bit wary to give too much in case it's homework, but I think it falls out pretty directly when considering the triangle $OPQ,$ from which we can find two sides and an angle. – Stephen Donovan Apr 19 '22 at 01:49
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    What have you tried to solve the problem? Add your efforts and attempts, that way, users can respond according to your level of understanding of the problem, not just give you the answer! – Clyde Kertzer Apr 19 '22 at 02:14
  • Thank you both for your comments. Just to add, this isn't for homework; I'm running physics simulations for a personal project; I'm sweeping $\theta$ and need to calculate the corresponding $\phi$ for each structure. I tried adapting the method described in the below post but the expression I got for $\theta$ was quite complicated and not correct. https://math.stackexchange.com/questions/599221/distance-from-point-in-circle-to-edge-of-circle – dogbiscuit123 Apr 19 '22 at 17:13

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As in @StephenDonovan comment, consider $\triangle OPQ $ and apply the law of sines

$ \dfrac{ \sin(\theta) } {r} = \dfrac{ \sin(\theta - \phi) }{r - a } $

From which it follows immediately that

$ \phi = \theta - \sin^{-1}\left( \dfrac{(r - a) \sin(\theta) }{r } \right) $

Hosam Hajeer
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  • There is another minor consideration, which is that $\sin(\theta - \phi) = x$ only necessarily implies $\theta - \phi = \arcsin(x)$ if $\theta - \phi$ is acute. Fortunately, this is easy to prove: consider $A$ and $B$ to be the points of intersection between circle $O$ and line $OP$ and by Thale's theorem we have that angle $AQB$ is a right angle, and clearly angle $OQP$ is less than it so it is acute. – Stephen Donovan Apr 19 '22 at 16:19
  • Thank you very much for this clear answer, very much appreciate it! – dogbiscuit123 Apr 19 '22 at 17:14