Stereographic projection of a disc. Stereographic projection maps circles of the unit sphere, which do not contain the north pole, to circles in the complex plane.
My question is what happens to the center of the circle in the complex plane, i mean what is the stereographic projection of the center?
More specifically, what does a disc in complex plane represent in the unit sphere(riemann sphere) i.e  stereographic projection of a disc in complex plane.
Does it represents a spherical arc?
 A: I guess by disc you mean a subset of the complex plane of the form $D_R(z_0)=\{z \in \Bbb C : |z-z_0|<R\}$ for some $z_0 \in \Bbb C$, $R>0$.
Now recall the pictorial description of the stereographic projection $\varphi$ of $S^2 \setminus \{(0,0,1)\}$ to $\Bbb C$, take an arbitrary point $w \in S^2 \setminus \{(0,0,1)\}$ and consider the straight-line segment (in $\Bbb R^3$) that starts at the 'North pole' i.e. $(0,0,1)$ and ends at your chosen $w$. One lets the line-segment extend and where it meets the Complex plane that is the image of $w$ under the stereographic projection.
Now consider a disc in $\Bbb C$ , say $D_R(z_0)$ for some $z_0 \in \Bbb C$, $R>0$.Then the boundary circle of this disc is $C=\{z \in \Bbb C : |z-z_0|=R\}$. Look at the pre-image of $C$ under the stereographic projection i.e. $\varphi^{-1}(C)$, it must be a circle on  $S^2$ (i.e. intersection of $S^2$ with a plane in $\Bbb R^3$) not containing the 'North pole' i.e. $(0,0,1)$. Then $S^2 \setminus \varphi^{-1}(C)$ is the disjoint union of two open sets $A_1$ and $A_2$, each of which are homeomorphic to hemispheres. Then exactly one of these $A_i's$ contain the North pole, say Without loss of generality $A_1$ . Now since the North pole is mapped to $\infty$ (the point at infinity of $\hat{\Bbb C}$) then  $\varphi(A_1)$ must be $\{z \in \Bbb C : |z-z_0|>R\}$ and thus $\varphi^{-1}(D_R(z_0))$ must be $A_2$ i.e. the open hemisphere that doesn't contain the North pole!
