Rolling of a circle along the positive $x$-axis without slipping and finding the locus of a point lying on the circumference of the circle. Consider the circle of radius $1$ with its centre at the point $(0,1)$. From this initial position, the circle is rolled along the positive $x$-axis without slipping. Find the locus of the point $P$ on the circumference of the circle which is on the origin at the initial position of the circle.
My work:
I found out the equation of the circle to be
$x^2+(y-1)^2=1$.
I have a weak feeling that the locus traced by the point $P$ might be helical.
But, I cannot do anything. Please help out, I don't have much idea about this type of problems.
 A: It is a bit unfortunate to use a unit circle-I will do it with a circle of radius $r$ which makes it more clear where things come from.  We can set $r=1$ at the end.  
When the circle has rotated through an angle $t$, the point of contact with the $x$ axis is $(rt,0)$.  The center of the circle is at $(rt,r)$  The angle to P, measured counterclockwise from a vector from the center of the circle and pointing right, is $\frac {3\pi} 2-t$, so the offset from the center of the circle to P is $(r\cos (\frac {3\pi}2-t),r\sin (\frac {3\pi}2-t))$, so the location of P is $(rt+r\cos (\frac {3\pi}2-t),r+r\sin (\frac {3\pi}2-t))$.  You can set $r$ to $1$ and use the angle-difference formulas to get rid of the $\frac {3\pi}2$
A: This is also known as the "curve of fastest descent". That link goes into great depth on the answer to the question 

If ... two points are vertically separated, what is the quickest way to get from between point A to point B?

The quick answer (as mentioned in previous comment above) is a cycloid. 
Part of a cycloid gives the path of fastest descent in use when designing rollercoasters, skate parks and in surfing etc.
