Sorting a permutation by reversals is a famous problem in genome rearrangements. Since 1997, quite some biological evidence were found that in many genomes the reversed regions are usually flanked by a pair of inverted repeats. This type of reversals are called symmetric reversals, which, unfortunately, were largely ignored until recently. In this paper, we investigate the problem of sorting by symmetric reversals, which requires a series of symmetric reversals to transform one chromosome $A$ into the another chromosome $B$. The decision problem of sorting by symmetric reversals is referred to as {\em SSR} (when the input chromosomes $A$ and $B$ are given, we use {\em SSR(A,B)}) and the corresponding optimization version (i.e., when the answer for {\em SSR(A,B)} is yes, using the minimum number of symmetric reversals to convert $A$ to $B$), is referred to as {\em SMSR(A,B)}. The main results of this paper are summarized as follows, where the input is a pair of chromosomes $A$ and $B$ with $n$ repeats. (1) We present an $O(n^2)$ time algorithm to solve the decision problem {\em SSR(A,B)}, i.e., determine whether a chromosome $A$ can be transformed into $B$ by a series of symmetric reversals. (2) We design an $O(n^2)$ time algorithm for a special 2-balanced case of {\em SMSR(A,B)}, where chromosomes $A$ and $B$ both have duplication number 2 and every repeat appears twice in different orientations in $A$ and $B$. (3) We show that SMSR is NP-hard even if the duplication number of the input chromosomes are at most 2, hence showing that the above positive optimization result is the best possible. As a by-product, we show that the \emph{minimum Steiner tree} problem on \emph{circle graphs} is NP-hard, settling the complexity status of a 38-year old open problem.