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A circle of radius $r$ with centre $C$ is located at distance $d$ from a point $P$.

There are two tangents to the circle which pass through point $P$ - one on each side. They intersect the circle at points $A$ and $B$.

What is the angle through $P$ between these two tangents? In other words, angle $APB$?

I know that angle $APB$ + angle $ACB$ add up to 180.

(Not homework, for graphics programming) Thanks, Louise

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  • $\begingroup$ What kind of graphics programming problem? $\endgroup$ – lightxbulb Jan 15 at 16:58
  • $\begingroup$ Recursive 2D radial tree layout with arbitrarily sized nodes! $\endgroup$ – Louise May 2 at 16:41
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Here is a picture:

enter image description here

$\overline {CP} = d$

$\angle CAP$ and $\angle BAP$ are right angles, and $\triangle APB$ is isosceles.

$m\angle APC = \arcsin \frac rd\\ m\angle APB = 2\arcsin \frac rd\\ m\angle BAP = \arccos \frac rd$

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  • $\begingroup$ Thank you! How did you generate the picture? $\endgroup$ – Louise Jan 16 at 17:14
  • $\begingroup$ I use MS Paint. $\endgroup$ – Doug M Jan 16 at 18:57
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I presume "at distance $d$" means that $|CP|=r+d$.

The triangle $ACP$ is right angled, with $|AC|=r$. Then $$\sin\angle APC=\frac{r}{r+d}.$$ Then $$\angle ABP=2\angle APC=2\sin^{-1}\frac r{r+d}.$$

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