<s>[INST] Find the distance between the points of two tangents along a circle I have the following problem: there is a circle with the $R = 5$ and center of the circle located on coordinate $(0, 0)$. I have two points $A(6, 8)$ and $B(-4, -6)$. From points, tangents to the circle were drawn. It is better illustrated as:

Let us denote points where tangents and circle intersect as $E, F, G, H$ (see the picture above for better the understandings). So we need to find the distance between E and F along the circle.
 [/INST] 
\begin{align}
|OE|=|OF|=
R&=5
,\quad
|OA|=10
,\quad 
|OB|=2\sqrt{13}
,\quad
|AB|=2\sqrt{74}
,\\
\triangle AOE:\quad
|AE|&=5\sqrt3
,\\
\triangle BFO:\quad
|BF|&=3\sqrt3
.
\end{align} 
\begin{align} 
\angle EOF&=\angle AOB-\angle AOE-\angle FOB
,
\end{align}
\begin{align} 
\angle AOB&=\arccos\frac{|OA|^2+|OB|^2-|AB|^2}{2\cdot|OA|\cdot|OB|}
=
\pi-\arccos(\tfrac{18}{65}\sqrt{13})
,\\
\angle AOE&=
\arccos\frac{|OE|}{|OA|}
=\tfrac\pi3
,\\
\angle FOB&=
\arccos\frac{|OF|}{|OB|}
=\arccos(\tfrac5{26}\sqrt{13})
,\\
\angle EOF&=
\tfrac{2\pi}3-\arccos(\tfrac{18}{65}\sqrt{13})
-\arccos(\tfrac5{26}\sqrt{13})
\approx 1.234262917
.
\end{align} 
So, 
the distance between $E$ and $F$ along the circle,
that is, the length of the arc $FE$ is
\begin{align} 
R\cdot\angle EOF&=
5\cdot(\tfrac{2\pi}3-\arccos(\tfrac{18}{65}\sqrt{13})
-\arccos(\tfrac5{26}\sqrt{13}))
\approx 6.171314600
.
\end{align}
Expression for $\angle EOF$ can be simplified to
\begin{align} 
\angle EOF&=
\arccos\frac{18+2\sqrt3}{65}
,
\end{align}
hence by the cosine rule 
we can also find 
\begin{align} 
|EF|&=\tfrac1{13}\sqrt{6110-260\sqrt3}
\approx 5.78698130
.
\end{align}
</s>