Finding length of arc intercepted by two tangent lines given radius and distance from circle

Question

I'm in the process of writing a script for work but realized that I need to calculate the length of a circular arc intersected by two tangent lines, and when I try finding it I get lost in a mess of trig that seems to go nowhere.

Here's what I have so far:

Tangent lines intersect each other at point $P$

Tangent lines intersect circle at points $A,B$

Circle radius is $R$

Point on circle nearest to $P$ is $Q$

Length of $\vec{PQ}$ is $T$

Using some trig, I found that $\vec{PA}=\vec{PB} = R\arccos(\frac{R}{R+T})$

But that's where I'm stuck. How can I find $\widehat{AQB}$ ?

Answer 1

$OAP$ is a right triangle because $PA$ is tangent to the circle. With other given data we are able to find angle $\widehat O$ in this triangle: $$\cos O = \frac{OA}{OP} = \frac{R}{R+T}$$ $$\widehat {AOQ} = arccos \frac{R}{R+T} $$ Now, it is easy to show that $\widehat{QOB}=\widehat{AOQ}$ and therefore $$\widehat {AOB} = 2 arccos \frac{R}{R+T}$$ The above is the angle of the arc $AQB$. If you are looking for the other arc that connects $A$ to $B$, it is $2 \pi - 2 arccos \frac{R}{R+T}$. And if you are looking for the angle $\widehat{AQB}$ , it is half of the latter arc: $$\widehat{AQB} = \pi - arccos \frac{R}{R+T} $$ .

May I ask, what kind of programming project are you involved in?

Answer 2

From $ \Delta AOP$ subtended angle is directly (since $OP$ is hypotenuse)

$$ 2 \cos^{-1}\dfrac{R}{R+T}$$

$R$ times the above is the length of the arc subtended by this angle.