In general, a power series of the form \sum_{n=0}^{\infty}c_{n}(x-a)^{n}=c_{0}+c_{1}(x-a)+c_{2}(x-a)^{2}+c_{3}(x-a)^{3}+\cdots∞∑n=0cn(x−a)n=c0+c1(x−a)+c2(x−a)2+c3(x−a)3+⋯ is said to be "centered at x=ax=a". The reason for this terminology is that a power series like this will converge on some interval centered at x=ax=a (though it might just converge at x=ax=a and it can also converge over the entire real line RR, which is not, technically, centered anywhere). In other words, it will converge on some interval with a at the midpoint of the interval (again, excluding the cases where it only converges at x=a or where it converges for all x\in RR).
For example, the real power series \sum_{n=0}^{\infty} 1/3^{n}(x-7)^{n}=1+1/3 (x-7)+ 1/9 (x-7)^2+ 1/27 (x-7)^3+\cdots is a geometric series and can be shown to converge for all x in the open interval (4,10). Note that this interval is centered at a=7.
Furthermore, in this last example, you can actually take x to be a complex number and the series converges for all x\in CC with the property that |x-7|<3 (the distance between x and 7 is less than 3), which is an open "disk" (the inside of a circle) in the complex plane centered at 7.
Taylor series are special kinds of power series. Given a function f that is infinitely differentiable over some open interval (c,d) containing the number x=a, you can compute the corresponding Taylor series for f centered at x=a as the following expression:
\sum_{n=0}^{\infty} f^{(n)}(a)/(n!)(x-a)^{n}=f(a)+f'(a)(x-a)+(f''(a))/(2!)(x-a)^2+(f'''(a))/(3!) (x-a)^3+\cdots
As before, this will converge on some interval centered at x=a.
For the geometric series example above, it happens to be the Taylor series for f(x)=3/(10-x) centered at a=7.
