Hi I am new to the university and yet now I face one problem I cannot solve! Pls help me x.x!

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Thanks :)

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- Sep 16th 2011, 10:37 AMNiosamaLength of the crossed belt.
Hi I am new to the university and yet now I face one problem I cannot solve! Pls help me x.x!

Attachment 22288

Thanks :) - Sep 16th 2011, 11:07 AMHallsofIvyRe: Length of the crossed belt! Pls Help!
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.

- Sep 16th 2011, 11:24 AMNiosamaRe: Length of the crossed belt! Pls Help!
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!