Proof that the radius of convergence exists
Yes the statement is correct. To see this: since $\sum a_n x_0^n$ converges so the sequence $(a_nx_0^n)$ is bounded say by $M$ (convergent to $0$) so
$$x_0\ne0\qquad |a_n x^n|=|a_n x_0^n|\left|\frac{x}{x_0}\right|^n\le M\left|\frac{x}{x_0}\right|^n$$ and for $0\le|x|<|x_0|$ the geometric series $\displaystyle \sum \left|\frac{x}{x_0}\right|^n$ is convergent.
I shall build upon the previous answers and try to prove the existence of a radius of convergence for the series, $$\sum a_n(z-z_0)^{n}$$ Let us assume two points, $ z_1$ and $z_2$, different from $z_0$ such that the series converges at $z_1$ and diverges at $z_2$. (This is the most general case, depending on the type of series, either of $z_1$ and $z_2$ may not exist). Thus we can define two sets, $A$ and $B$, given by: $$ A := \{|z-z_0|: \sum a_n(z-z_0)^n\hspace{0.1 cm} converges\}$$ $$ B := \{|z-z_0|: \sum a_n(z-z_0)^n\hspace{0.1 cm} diverges\}$$
Now using the theorem that the users have mentioned before, we can conclude:
$ \sum a_n(z-z_0)^n$ diverges at z for $|z-z_0|>|z_2-z_0|$, thus$|z_2-z_0| \geq r$ for all $r \in A$, and using the completeness property of real numbers, $A$ being a non empty bounded above set, should have a supremum, $r_1=sup(A)$.
Also, $ \sum a_n(z-z_0)^n$ converges at z for $|z-z_0|<|z_1-z_0|$, thus$|z_1-z_0| \leq r'$ for all $r' \in B$, and using the completeness property of real numbers, $B$ being a non empty bounded below set, should have n infimum, $r_2=inf(B)$.
It just remains for us t show that $r_1=r_2$, and we shall be done.
If possible we let $r_1\neq r_2$, using law of trichotomy, we should either have:
$$r_1>r_2$$ Clearly we can chose a $u\in \mathbb{R}$ such that $r_1>u>r_2$, now that leads to trouble. Let $z^{*}\in \mathbb{C}$, such that $|z^{*}-z_0|=u$. This indicates that for the above z, the series must converge as well as diverge. (Using the facts that $ \sum a_n(z-z_0)^n$ diverges at z for $|z-z_0|>|z_2-z_0|$ and $ \sum a_n(z-z_0)^n$ converges at z for $|z-z_0|<|z_1-z_0|$)
$$r_1<r_2$$ Cleary there exists a $u \in \mathbb{R}$ such that $u\in (r_1, r_2)$, and $u\notin A$ and $u\notin B$, which is impossible as for $z$, such that $|z-z_0|=u$, the series neither diverges, nor converges, which again, is a contradiction.
Thus we arrive at the inevitable conclusion that $r_1=r_2=r$(say), and using the properties of r, it is obvious that:
$\sum a_n(z-z_0)^n$ converges for $|z-z_0|<r$ and diverges for $|z-z_0|>r$