To find that shape and length

Question

A circle of radius 'r' is centered at origin, consider an imaginary line y=r (tangent to the circle at 0,r say A)

Now what kind of shape should an arc (not necessarily circular) have that if initially it's one end is at A and being tangential to circle at A, then on rolling the arc over the circle (just like tyre rolls without sliding over road) the 'A' end of the arc always traces/'have locus on' y=r

Answer 1

Your question is not very clear at all.

In the differential geometry of curves, a roulette is a kind of curve that is described by a point (called the generator or pole) attached to a given curve and that curve rolls without slipping, along a second given curve that is fixed.

See https://en.wikipedia.org/wiki/Roulette_(curve)

We have :

• cycloids, https://en.wikipedia.org/wiki/Cycloid

• epicycloids, https://en.wikipedia.org/wiki/Epicycloid

• hypocycloids, https://en.wikipedia.org/wiki/Hypocycloid

• trochoids, https://en.wikipedia.org/wiki/Trochoid

• involutes, https://en.wikipedia.org/wiki/Involute.

As you are working with "a circle of radius 'r' is centered at origin", we have to deal with epicycloids, hypocycloids or involutes.

Because of your english, I suspect you are referring to the involute.

But please, confirm.

Answer 2

After the comments mantained with Anshuman Dwivedi, I am posting a new answer.

**

I am solving the following problem:

A circle of radius $r$ is centered at origin.

Consider an imaginary line $y=r$, tangent to the circle at A = (0,r), and a segment AB of length $l$ on this line.

What shape will this segment describe if it moves rolling around the circumference and maintaining its point A on line $y=r$.

For any position of the segment around the circumference, we have:

We advance point A and rotate the segment till it is tangent to the circumference.

Coordinates of tangential point are calculated as follow

$Alfa= 2*atan((x-coordinate-of-point-A) / r)$

$Tx = r * cos( \pi/2- Alfa)$

$Ty = r * sin( \pi/2- Alfa)$

Now, new position of point B will be

$Bx = ( x-coordinate-of-point-A ) + l * cos(Alfa)$

$By = r - l*sin(Alfa)$

Following is the curve obtained

https://i.stack.imgur.com/QxRPJ.jpg