infact no. of such circles so called halving circles is n^2 .this is due to federico ardilla.

who wrote: infact no. of such circles so called halving circles is n^2 .this is due to federico ardilla. Do you have a proof of this fact? Please give a link of sth?

http://arxiv.org/abs/math/0408354v1

I believe there is also a solution using inversion as Professor Andreescu once gave.

In the article mentioned above, Ardila says about this problem (which he proposed): "[It] follows easily from the nontrivial observation that, for any $A$ and $B$ in $S$, the number of halving circles that go through $A$ and $B$ is odd." Indeed, if that is the case, then we can work mod 2, so that the total number of such circles is $\frac{1}{3}\binom{2n+1}{2}\pmod{2}$ which is equal to $n\pmod{2}$. Now we prove the claim made in the first paragraph. I will rename the pair of points in $S$ to $P$ and $Q$, since that is what the source used I am copying from. Divide the $2n-1$ other points into two groups depending on which side of $PQ$ they lie. By the condition that no three points are collinear, each of these points fall into one of these groups. Label the points in these groups $A_1,A_2,\ldots,A_m$ and $B_1,\ldots,B_p$ respectively, where $m+p=2n-1$. Let the angles $\angle PA_1B, \ldots, \angle PA_mQ$ be $\alpha_1,\ldots,\alpha_m$, respectively and the angles $\angle PB_1Q, \ldots, PB_pQ$ be $\pi-\beta_1,\ldots, \pi-\beta_p$, respectively. Without loss of generality we assume that $\alpha_1>\alpha_2>\cdots>\alpha_m$ and $\beta_1>\beta_2>\cdots>\beta_p$. Note that none of these angles can be equal, otherwise we would have a circle passing through four of the points. No order all the angles $\alpha_1,\ldots,\alpha_m,\beta_1,\ldots,\beta_p$ from largest to smallest. The circle through $P$, $Q$ and $A_1$ contains the point $A_k$ (in its interior) if and only if $\alpha_k>\alpha_i$ if and only if $k<i$. Similarly $B_l$ lies in the interior of circle $PQB_j$ if and only if $l>j$. Also, $B_j$ lies in the interior of circle $PQA_i$ if and only if $\alpha_i+\pi-\beta_j>\pi$ if and only if $\alpha_i>\beta_j$. Similarly, $A_i$ lies in the interior of circle $PQB_j$ if and only if $\alpha_i>\beta_j$. Now consider and ordering of all the angles and suppose that there is an instance where $\cdots>\alpha_i>\beta_j>\cdots$ are consecutive in this ordering. We claim that circles $PQA_i$ and $PQB_j$ have the same number of interior points. Indeed, circle $PQA_i$ contains the points $A_1,\ldots,A_{i-1}$ and $B_j,\ldots, B_p$, while circle $PQB_j$ contains the points $A_1,\ldots, A_i$ and $B_{j+1},\ldots, B_p$. Consider the following transformation of $S$: Replace $A_i$ by another point $A'_i$ on the same side of $PQ$ such that $\alpha'_i:=\angle PA'_iQ=\beta_j$ and replace $B_j$ by another points $B'_j$ on the same side of $PQ$ such that $\pi-\beta'_j=\angle PB'_jQ=\pi-\alpha_i$. The only change in the total ordering is that $\alpha'_i$ and $\beta'_j$ are in the opposite order to $\alpha_i$ and $\beta_j$. So now we have $\cdots>\beta'_j>\alpha'_i>\cdots$. When $k\ne i$ and $l\ne j$, performing this transformation does not change the number of points in the interior of circle $PQA_k$ or $PQB_l$. Thus the only circles that may change from good to bad or bad to good are $PQA_i$ and $PQB_j$. Before the transformation circles $PQA_i$ and $PQB_j$ contained the same number of interior points, and thus either both were good or both were bad. A similar counting shows that circles $PQA'_i$ and $PQB'_j$ have the same number of interior points after the transformation, and thus also are either both good or both bad. So performing this transformation changes the number of good circles by either $+2$ or $-2$, and thus preserves the parity. Now we repeat this transformation until we end up with a configuration where $\beta'_1>\beta'_2>\cdots>\beta'_p>\alpha'_1>\cdots>\alpha'_m$. For each $i$ the circle $PQA_i$ contains the $i-1$ points $A_1,\ldots, A_{i-1}$, while for each $j$, the circle $PQB_j$ contains the $p-j$ points $B'_{j+1},\ldots, B'_p$. So, if $m\ge n$, then there is exactly one good circle $PQA_i$, namely when $i=n$; and if $p\ge n$, there is exactly one good circle $PQB_j$, namely when $j=p+1-n$. Since $m+p=2n-1$, exactly one of the relations $m\ge n$ or $p\ge n$ can hold, and therefore this arrangement has exactly one good circle.