As I already said, I am young and I never studied vectors, I have no idea what normalization is.
Normalization is a process where you take a vector different than zero and create another vector that has the same direction but has a length of 1.
Let v = (x,y) be any vector different than (0,0).
length(v) = sqrt(x2+y2)
Now, define w = v/length(v) = (x/sqrt(x2+y2), y/sqrt(x2+y2))
If you want the length of "w" then:
length(w) = sqrt( ((x/sqrt(x2+y2))2 + (y/sqrt(x2+y2))2 )
length(w) = sqrt( x2/(x2 + y2) + y2/(x2 + y2))
length(w) = sqrt( (x2+y2)/(x2+y2) )
length(w) = sqrt(1)
length(w) = 1
One of the advantages of normalizing a vector is that when you multiply a vector of length 1 by a single value, that value becomes the length of the new vector.
So, if length(w) = 1 and you define z = r * w (r a single value, not a vector), then length(z) = r.
Another thing that's usually done in algebra is using the equations around the origin, not a random point, so it's easier to move the reference of your problem to the origin, solve the problem, and translate the solution back to the point you want.
All this is usefull for your problem.
First, translating the problem, you what a point that's at a distance of "r" from "p", so you can translate the whole problem to the origin by substracting "p". Your point "t" becomes "t-p" in this case.
Second, you use the normalization to make a vector of length 1 form the origin to "t-p", then multiply it by "r" to get a point in the circle of radius "r". That point is the solution around the origin, so you add "p" to it to move the solution back to your original "p" point.