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.
\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}
If we pick point $E$ for example, then we know that $EO \perp EA$. This means that $(\text{slope of EO})(\text{slope of EA})$ equals $-1$. If $E = (x, y)$, we have:
$$\frac{8-y}{6-x} \cdot \frac{y - 0}{x - 0} = -1$$
$E$ also lies on the circle, so:
$$x^2+y^2=5^2$$
which gives two possibilities for $(x,y)$, where one is $E$ and the other is point $G$.
You can repeat this same process to find $F$, after which you can just use the distance formula.
Note that the chord of the circle $x^2+y^2=r^2$ passing through the two tangent points drawn from the external point $(x_1,y_1)$ is $x_1x+y_1y =r^2$. Then, the equations of the chords EG and FH are, respectively
$$6x+8y=25,\>\>\>\>\>-4x-6y=25$$
Substitute them into $x^2+y^2=25$ to obtain the points $E(\frac32+2\sqrt3, 2-\frac32\sqrt3)$ and $F(\frac{45}{26}\sqrt3-\frac{25}{13}, -\frac{75}{26}-\frac{15}{13}\sqrt3)$, which yields the distance
$$EF=\sqrt{\frac{10}{13}(47-2\sqrt3)}$$
