I am a software developer trying to remember relevant terminology and math formulas that I haven't used in well over a decade. As such, this question is trivial and likely has a solution somewhere on the internet but I don't know what to search to find the answer! Here's the problem:
Given a circle with radius $r$, the angle $\alpha$ can be described by the point
$P_{(x,y)} = \cos(\alpha) \cdot r, \sin(\alpha) \cdot r$
We then define the point
$S_{(x,y)} = \cos(\alpha)\cdot r,\sin(\alpha)\cdot r +d$
where $d$ is some value.
Point $R$ is then defined as intersection of the horizontal line drawn from $S$ to intersect the circle (selecting the smaller of the two possible line-segments assuming $\alpha$ is not $\frac{\pi}{2}$ or $\frac{3\pi}{2}$).
What is a formula to define the angle $\beta$ between points $P$ and $R$ and the center of the circle?
Thank you!
Best Answer
The formula should be, at least for the first quadrant, $0 \leq \alpha < \pi/2$:
$$\beta=2 \ atan \left( \dfrac{e}{f}\right) \ \ (1)$$
where $e$ is the length of SR i.e.,
$$e=\sqrt{r^2-(r \sin \alpha - d)^2}-r \cos \alpha \ \ (2)$$
and
$$f=2 r \sin \alpha - d \ \ (3)$$
Why all that ? Let $P'$ be the symmetrical point $P'$ of $P$ with respect to x-axis, and $e=SR$ as said before.
the coordinates of $R$ being $(r \cos \alpha + e, r \sin \alpha -d)$, we have to express that $R$ belongs to the circle with radius $r$ by writing $(r \cos \alpha + e)^2+( r \sin \alpha -d)^2=r^2$, out of which we get (2).
the fact that $PP'=2 r\sin \alpha$ is clear, thus $P'S=2 r\sin \alpha -d$ whence (3).
the central angle theorem (http://www.mathopenref.com/arccentralangletheorem.html) is the last ingredient. Explanation: the angle from which arc PR is seen from point $P'$ is $\beta/2$, the half of angle $\beta$ from which the same arc PR is seen from the center of the circle. This angle $\beta/2$ is such that, in triangle $P'SR$, $\tan(\beta/2)=SR/SP'=e/f$, explaining formula (1).