Does tangential acceleration change with radius? Do tangential velocity and tangential acceleration change with radius (change of radius on the same object)? 
For example consider a spinning disk. Does the equation $$a_t = \alpha R$$ (where $a_t$ is the tangential acceleration, $\alpha$ is the angular acceleration and $R$ is the radius of the disk) give me the tangential acceleration of the centre of mass or of a point on the edge of the disk? 
As you go inwards from a point on the edge of the disk , the radius decreases. So doesn’t that mean the tangential acceleration of the centre of mass is zero? I have the same doubt regarding tangential velocity. What is wrong with my reasoning? Does the centre of mass have the highest tangential velocity and acceleration or the lowest of all points on the disk?
 A: To an extent, you are correct in your assumptions.
The formula $$\mathbf{\vec a = \vec \alpha \times \vec R \implies  a=R \, \alpha }$$ when $\vec \alpha$ and $\vec R$ are perpendicular to each other, which is quite the general case.
The aforementioned formula relates the tangential acceleration $\vec a$ of a point particle placed at a distance $R$ from the center of rotation to the angular acceleration $\vec \alpha$ at that point. And if the rotating center has no translatory motion, then the tangential acceleration described by the above equation is equal to the net linear acceleration of the particle.
And the same is true for the tangential velocity as well, which goes as:
$$\mathbf{\vec v = \vec \omega \times \vec R \implies  v=R \, \omega }$$
So, if $R$ decreases in magnitude, that is, you move to points which are closer to the rotation center, then obviously you get lower values of tangential velocity as well as tangential acceleration.
And, at the center of rotation, both the tangential velocity and tangential acceleration are zero, at all times.
Hope this helps.
A: A note on the case you're considering:
A disk is an extended body. This means it's a collection of points: and must be treated as such. To speak of the displacement/velocity/acceleration of a point on the disk and of the disk itself - the center of mass of the disk - are two distinct analyses.

Particles on the disk
Particles of the disk all travel in circles around the axis of rotation. (From your question, it's implied that the axis passes through the center of mass, so we'll continue to use the same convention (call the axis $x$).)
First, any given point is a part of two disjoint sets: it's either part of the disk, or it isn't. Since the disk is rotating, any particle that's on the disk is in circular motion around $x$. If it isn't part of the disk, it's not in circular motion.
Clearly, any particle not on the disk has zero angular velocity. Of the points on the disk,  all those with non-zero distance from the axis have some tangential velocity.
$$\vec v = \vec\omega \times\vec R$$
Since $x$ passes through the centre of mass, its tangential velocity is, as correctly stated, zero.
A: Firstly defining circular motion 
Circular motion is when a body moves in circle or as they say have a fixed distance from a point (moving or stationary)
Now when we consider rotating rigid bodies.We usually take torques ,angular momentum, moment of interia about the point which is stationary with respect to ground. As a disc spins freely and doesnt perform pure roll,the following qualities will be account about the center of the disc.
Okay now you may ask why?
The formula for angular velocity is a vector equation
Mod(v1 vector-v2 vector)/distance between the point 1 and 2
As velocity of center is zero center is considered
If you wish to take any other point use this vector equation.
Same is valid for angular velocity is you differenciate the equations with respect to time
