To get the electric potential at a certain point due to a uniform circular disc of charge, you first have to look at the contribution to the potential from each infinitesimal area unit of the disc at that point.
I find that, for objects such as this disc which have circular symmetry, it is easiest to deal with polar coordinates (in two dimensions, in this case). In polar coordinates, where a point is defined by its distance r from the origin and its azimuthal angle θ (typically measured counterclockwise from the +x axis), a unit of area is defined by (rdrdθ), as long as θ is measured in radians. This area in Cartesian coordinates would have been dxdy.
So each infinitesimal point on the disc has a charge equal to the uniform charge per unit area p times the area rdrdθ. In other words, each infinitesimal bit of area has charge dq = prdrdθ.
We can calculate the electric potential a distance d from this infinitesimal unit of charge using the expression for electric potential of a point charge.
V = kq/d --> V = k(prdrdθ)/d for the infinitesimal unit of charge we are dealing with.
Then, to find the total electric potential at some point, you need to add up all the contributions to the potential from every infinitesimal unit of charge on the disc. This will involve integration, but, before that, we have to figure out how to write d in general to the desired point from every point on the disc.
This will be tricky without diagrams, but let me see if I can set it up.
The point at the center is actually relatively straightforward. The distance from any point on the disc to the center is just r, which is the distance coordinate in polar coordinates. So the contribution to the potential at the center from an infinitesimal point on the disc at a distance r and angle θ (in polar coordinates) is:
dV = kdq/r = kprdrdθ/r = kpdrdθ
Then, to get the total potential at the center from the entire disc, integrate this expression across the disc. That is, from 0 to a in the r direction, and from 0 to 2π in the θ direction (or you could just integrate from 0 to π in the θ direction and multiply by 2, because of the symmetry; it gets the same result).
So the integral is:
V = ∫0a ∫02π kpdrdθ
With k and p being constants (k is the Coulomb constant), this is a pretty straightforward integral. If you want to proceed further with this or check an answer, just let me know.
Note that, because electric potential is a scalar, we do not have to worry about directions the way we do when determining, for example, the net electric field at a given point from a charge distribution.
Determining the electric potential at a point on the circumference is a trickier beast. The overall process is the same, but determining an expression for the distance from any point on the disc to a point on the circumference is a little more involved, since it does not benefit from the nice symmetry that the point at the center has (always being a distance r from any point on the disc). This is why you often see the problem solved for a point along the axis of symmetry, instead. Physicists like symmetry :).
But, in this case, we do not have it, so we have to work another way. Here is where a diagram would really help. But, failing that, I can say that the best way I found to generalize the distance from any point on the disc to a point on the circumference, using polar coordinates, was through the law of cosines.
Take an infinitesimal point at polar coordinates r and θ. This infinitesimal point still has charge dq = kprdrdθ. What is its distance d from a point on the perimeter? You can draw a triangle where the vertices are the center of the circle, the infinitesimal point of charge at (r,θ) and the point on the perimeter. The side of the triangle from the center to the infinitesimal point at (r,θ) is r. The length of the side of the triangle from the center to the point on the perimeter is a. The length of the third side is the distance from the infinitesimal point of charge to the point on the perimeter. This is the length d we want an expression for. The angle opposite the side d is the angle θ in the coordinate of the infinitesimal point of charge.
If you can picture that, then you may be able to see that the law of cosines relates the side length d to the other two side lengths and the angle θ, as:
d2 = r2+a2-2racosθ --> d = (r2+a2-2racosθ)1/2
Thus, we have, for the contribution to the potential on the perimeter from an infinitesimal point of charge at (r,θ):
dV = kdq/d = kprdrdθ/(r2+a2-2racosθ)1/2
To get the total potential at that point on the perimeter, this expression must be integrated, as before, over the entire disc:
V = ∫0a ∫02π kprdrdθ/(r2+a2-2racosθ)1/2
This is a less appealing integral than the last one (which is why findiing the potential on the perimeter of a uniform charge distribution is not as common a problem as finding it for positions with more symmetry). I will not work it out yet, but we can look into that more later, if you like. I would suggest integrating θ first, since -- through the chain riule -- it basically boils down to integrating dθ/(a+bcosθ)1/2, which may have a standard table solution or be not too bad to work out.
Or perhaps someone else has a simpler way to set up the problem. But this is all I can see for this part right now.
As I said, if you want to talk more about the integration, or how this was set up, please let me know. I know it is not easy to follow sometimes without being able to draw. But I hope this helps you move along with the problem!