Length of a belt? I know this is vague, but I am studying for a test. I remember the shape of the object, but not much else.
Basically I have to find the length of a belt, when given the radius of a circle. 
The belt goes around the circle and to the point. I did not give a radius, but you can make it up. I just cannot remember how to find the length of the belt.
 A: Hint:  Draw radii to the points of tangency and a line from the center to the kink.  The radii are perpendicular to the tangent line, so you have a pair of right triangles.  If you have an angle, you can use trig to get the length of the tangent.  You can also use the right triangles to determine how much of the circumference is covered by the belt.
A: Let $r$ be the radius of the circle and $a$ be the distance from the center of the circle to the "kink".  The central angle is $\theta = \cos^{-1} (\frac{r}{a})$.  The length of the part of the belt that wraps around the circle is then $2 \pi - 2r \theta$.  The length of the part of the belt that goes from the point of tangency to the kink and back to the circle at the other point of tangency is $2a \sin{\theta}$.  Thus the length of the belt is $2 \pi - 2r \theta + 2a \sin{\theta}$. There is an algebraic formula you could use for $\sin{\theta}$.
A: Let the radius of the circle be $R$ & the distance of the given point (kink) say $P$ from the center say $O$ of the circle be $d$ then draw two tangents from the point $P$ to the circles at the points say $M$ & $N$. Thus we have
Length of each of the straight portions of the belt $$PM=PN=\sqrt{(OP)^2-(OM)^2}=\sqrt{d^2-R^2}$$ 
Now, the aperture angle say $\theta$ subtended by the minor arc $MN$ at the center $O$ of the circle $$\cos\frac{\theta}{2}=\frac{OM}{OP}=\frac{R}{d}$$ $$\implies \theta=2\cos^{-1}\left(\frac{R}{d}\right)$$ Length of the curved portion of the belt i.e. length of major arc $MN$ $$=(2\pi-\theta)R=\left(2\pi-2\cos^{-1}\left(\frac{R}{d}\right)\right)R$$ $$=2\left(\pi-\cos^{-1}\left(\frac{R}{d}\right)\right)R$$
Hence, length of the belt $$=PM+PN+\text{major arc}\space MN$$  $$=\sqrt{d^2-R^2}+\sqrt{d^2-R^2}+2\left(\pi-\cos^{-1}\left(\frac{R}{d}\right)\right)R$$ $$=\color{blue}{2\left[\pi R+\sqrt{d^2-R^2}-R\cos^{-1}\frac{R}{d}\right]}$$
