Do you see that the center of the two circles is at (2, 2) and (10, 8)? And since they have the same radius, the situation is symmetric: the belts cross at $\displaystyle \left(\frac{2+ 10}{2}, \frac{2+ 8}{2}\right)= \left(6, 5)$. If (a, b) is the point where the first touches the second circle, the slope of the line from (6, 5) to (a, b) is $\displaystyle \frac{b-5}{a- 6}$. Also, the slope of the line from the center of that circle to (a, b) is $\displaystyle \frac{b-8}{10- a}$. Since those two lines must be perpendicular, it must be true that their product is -1: $\displaystyle \frac{b-5}{a- 6}= \frac{b- 8}{10- a}$. That gives you one equation in a and b. Since (a, b) is on the second circle, you also have $\displaystyle (a- 10)^2+ (b- 8)^2= 9$. You can solve those two equations for a and b.
But How do I find the length! I think I need to find errr The Length of each side of the two triangles and since the belt is around for other half that mean I need to find half length of each circle. I have found the half length of the two circle, distance from center point of two circle. One side from each triangle which are both 6. I still need other side so I can sum up! I cannot figure that out!