Find Coordinates of the Center of a Circle After It's Made a Quarter-Turn The question is below.
I was able to solve part (a) because the $x$-coordinate would just be the circumference of the circle, which is $2\pi$. Therefore, $P = (2\pi, 0)$. 
I am confused with parts (b) and after,  but I know that the $y$-coordinate will remain to be $1$ because that's the radius of the circle and how far it is off from the $x$-axis. Any help with how to find the $x$-coordinate will help for part (b). Thank you in advance. 
 A: The centre point travels 2πr for every complete turn along x. Similarly it travels (1/4)2πr for a quarter turn.
Then turn the figure vertically with the x axis pointing down. Imagine the angle of reference as the second hand of a clock turning clockwise, starting at 9 o’clock. This means the y coordinate of the point is (the absolute value of) the sine of the angle, scaled by the radius (take another look at the definition of sine in the unit circle). In this case, after a quarter turn the y coordinate is r⋅sin((1/4)2π)
A: The $x$-coordinate is just the length of the "rotated arc", i.e., $\displaystyle \frac{\pi}{2}$.
A: The parametric equation of point $P$ of cycloid on rolling wheel circle radius $a=1$
$$ x= a (\theta- \sin \theta),\quad y= a(1-\cos \theta) $$
With a bit of calculus horizontal slope is 
$$ \frac{dy/d \theta}{dx/d \theta}= \tan \phi_h = \cot \phi =\frac {\sin \theta }{1- \cos \theta}= \cot \frac{\theta}{2} \rightarrow  \boxed{\theta = 2 \phi} $$
So  amount crank radius rotates is double of what the cycloid vertical tangent tardily rotates.

This parametric form I think answers all your questions.
The y-cordinate of center is always $a$ no matter what the amount of roll $\theta $ is. The question is designed to confuse you,  it appears.
