The length of an arc of a circle is 12

Today, I ran into a calculus class and the professor was calculating the perimeter of a circle using arc lenghts. Even though they seemed to be on the right track, there is a problem:

They are using sinx and cosx functions, which are actually defined using the perimeter of a circle, aren't they? So, in my opinion, that was not a real proof since they were using what they want to show.

Or am I missing something? Was that a real proof?

Edit: I don't know but most probably they defined sinx and cosx in the usual way, because it was a calculus class.

How to find the center of a circle with given an arbitrary arc. we only have the arc nothing else. Is there any known equation or way to complete the circle.

It is not possible for a part of any of three conic sections to be an arc of a circle.
It is asked to prove this theorem without using any notation or any modern form of symbolism whatsoever (algebra). But, to me, this seems to be quite a difficult task. How can this be done?

Suppose I'm given two points: $(x_1,y_1)$ and $(x_2,y_2)$ (which are real numbers) lying on the circumference of a circle with radius r and centred at the origin, how do I find the arc length between those two points (the arc with shorter length)?

Let $n$ points be placed uniformly at random on the boundary of a circle of circumference 1.

These $n$ points divide the circle into $n$ arcs.

Let $Z_i$ for $1\le i\le n$ be the length of these arcs in some arbitrary order, and let $X$ be the number of $Z_i$ that are at least $\frac{1}{n}$.

What is $E[X]$ and $Var[X]$?

Any hints will be appreciated. Thanks..

(By the way this problem is exercise 8.12 from the book 'Probability and Computing' by Mitzenmacher and Upfal)

What angle is subtended at the center of a circle of radius 2 km by an arc of length 9 m?

I am trying to figure out what the best method is to go about finding this locus.
$\arg {\displaystyle \frac{z-a}{z-b}}=\theta$
I am aware that it must be part of an arc of a circle that passes through the points $a$ and $b$. The argument means that the vector $(z-a)$ leads the vector $(z-b)$ by $\theta$ and so by the argument must lie on an arc of a circle due to the converse of 'angles in the same segment theorem'.

My question is, how do I know what the circle will look like, ie. the centre of the circle and the radius. If I don't need to know this, how will I know what the circle looks like.

Finally, if I changed the locus to
$\arg {\displaystyle \frac{z-a}{z-b}}=-\theta$
Would this just be the other part of the same circle as previously or a different circle?

Perhaps you could help by showing my how it would work with the question
$\arg \frac{z-2j}{z+3}=\pi /3$