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

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

• What kind of graphics programming problem? – lightxbulb Jan 15 at 16:58
• Recursive 2D radial tree layout with arbitrarily sized nodes! – Louise May 2 at 16:41

Here is a picture:

$$\overline {CP} = d$$

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

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

• Thank you! How did you generate the picture? – Louise Jan 16 at 17:14
• I use MS Paint. – Doug M Jan 16 at 18:57

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