What is the angle between two intersecting tangents to a circle?

Question

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

Recursive 2D radial tree layout with arbitrarily sized nodes! — Louise, Commented May 2, 2019 at 16:41

Doug M · Accepted Answer · 2019-01-15 17:58:15Z

1

Here is a picture:

$\overline {CP} = d$

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

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

answered Jan 15, 2019 at 17:58

Doug M

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$\begingroup$ Thank you! How did you generate the picture? $\endgroup$
– Louise
Commented Jan 16, 2019 at 17:14
$\begingroup$ I use MS Paint. $\endgroup$
– Doug M
Commented Jan 16, 2019 at 18:57

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Angina Seng · Accepted Answer · 2019-01-15 16:59:07Z

0

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 $$\angle ABP=2\angle APC=2\sin^{-1}\frac r{r+d}.$$

answered Jan 15, 2019 at 16:59

Angina Seng

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