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I've been searching the internet all day for this particular case and can't find a single word on the subject - whether that's my poor searching or the subject is obscure, I do not know. I am also not particularly fluent with mathematical jargon, so forgive my fumbling explanation. Hopefully one of you may re-phrase my problem after I present it.

This Tangent Chord Angle diagram is the closest visual representation I can find for describing my dilemma. I'll be referring to points from this diagram, although it doesn't demonstrate the problem itself, only the starting conditions of my problem.

I have been trying to write a program that will rotate a vector TP from a point T tangent to circle O until that vector essentially describes a chord on the circle (possibly TA if the magnitude of A equals the magnitude of P).

I know:

  1. T, the initial tangent point
  2. TP, the initial vector from T
  3. O, the circle on which T lies
  4. The radius of circle O

I have no clue what to do to get that vector to "fit" inside the circle. I understand there are two possible chords depending on the direction of rotation - I'll be using whichever angle is closer to the vector's original angle.

Again, forgive my inability. I'm afraid it would take considerably longer for me just to learn how to more appropriately describe the problem.

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As it turns out, the tangent-chord angle is actually just what you want. But you want the particular tangent-chord angle for the chord that is the same length as your vector. That is, you want to find the angle $\angle ATP$, where a point $A$ on the circle $O$ such that $\lvert TA\rvert = \lvert TP\rvert$ (that is, the lengths of the vectors from $T$ to $P$ and to $A$ are equal). Moreover, of the two possible solutions, you want the one that gives the smaller value of $\angle ATP$.

Then since $T$ and $A$ are both on the circle, they are at equal distances from $O$, and $\triangle AOT$ is an isoceles triangle with side lengths $\lvert OA\rvert=\lvert OT\rvert=r$, where $r$ is the radius of the circle. Let $M$ be the midpoint of $TA$; then $\triangle OMT$ and $\triangle OMA$ are two congruent right triangles with hypotenuse $r$ and leg length $\frac12 \lvert TA\rvert$ opposite the angle at $O$, and $$\angle AOT = \angle AOM + \angle MOT = 2\angle MOT.$$

But by the tangent-chord angle theorem, $\angle ATP = \frac12\angle AOT$, so $\angle ATP = \angle MOT$. And by the definition of the sine of an angle, $$\sin(\angle MOT) = \frac{\frac12 \lvert TA\rvert}{r}.$$ Noticing that $$\frac{\frac12 \lvert TA\rvert}{r} = \frac{\lvert TA\rvert}{2r} = \frac{\lvert TP\rvert}{2r},$$ we find that $$\sin(\angle ATP) = \sin(\angle MOT) = \frac{\lvert TP\rvert}{2r}.$$

You know $\lvert TP\rvert$ and $r$, so you just need to solve for $\angle ATP$ in the equation above. The solution is $$ \angle ATP = \arcsin\left( \frac{\lvert TP\rvert}{2r} \right); $$ of the possible angles that would solve the equation for $\sin(\angle ATP)$, this chooses the smallest angle, which is the one you want. The mathematical function $\arcsin$ is implemented as asin in many software libraries; the output of this function is usually an angle measured in radians.

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  • $\begingroup$ But does the tangent-chord angle theorem assume that TP is itself tangent to the circle? I only know that T is a tangent point. The usual case for my program is that TP is actually just a continuation of OT on that same angle, but is frequently offset in one direction or another. I did say that the diagram most closely represented my problem, but I don't think it's 100% accurate. I'm sorry, I did a poor job trying to lay out the entirety of the situation. Would the same solution apply? $\endgroup$ – JRHard Dec 6 '16 at 16:57
  • $\begingroup$ The tangent-chord angle theorem does assume that $TP$ is tangent to the circle. If $TP$ is not tangent to the circle, then the amount of rotation is whatever it takes to rotate $TP$ to the tangent direction at $T$ and then rotate it $\arcsin(\lvert TP\rvert /(2r))$ toward the inside of the circle. Another way to look at it is that $\angle OTA =\arccos(\lvert TP\rvert /(2r))$; the rotation is whatever would rotate $TP$ to point directly at the center of the circle, minus $\angle OTA$. $\endgroup$ – David K Dec 6 '16 at 20:07

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