about shape of open ball in metric space consider following shape in plane :

now we have definition of open ball in every metric space :
$$B_r(x_0) := \{x \in X : d(x_0,x)<r\}$$
that radius is $r$ and center is  $x_0$ , my questions are :
1-which one above shapes can be shape for open ball in a metric space ? (or what are sufficient conditions for every shape that must be open ball ? ) 
2-which one above shapes are equivalent topologically ?
3-what are properties of open ball ? fo example every open ball is convex set .
4- can we find in every shape of open ball 'center and radius' ?
I think shape 3 can not be a open ball because it is not convex.
and shapes 1,4 and 7 are equivalent topologically because with defintion of equivalent topologically with base in munkres book we can consider following shape:
(note that we have $B_{r,shape 7}(x_0) \subseteq B_{r,shape 4}(x_0) \subseteq B_{r,shape 1}(x_0) $ )


$\bullet$Example 1. The discrete metric: $d(p,q)=1$ if $p\ne q,$ and $d(p,p)=0$, on any set $S.$ Every subset of $S$ is open. We have $B_d(p,r)=\{p\}$ if $0<r\leq 1.$ And $B_d(p,r)=S$ if $r>1.$ $\bullet$Example 2. The river metric on $R^2$ defined by  $d(\;(x_1,y_1),(x_2,y_2)\;)=|x_1-x_2|+|y_1|+|y_2|$ if $x_1\ne x_2|$, and $d(\;(x,y_1),(x,y_2)\;)=|y_1-y_2|.$ Lines parallel to the $y$-axis are separated by mountains. To go from one to another, you have to go to the river (the $x$-axis) to travel horizontally. An open ball $B_d(\;(x,0),r)$ is $\{(u,v):|u-x|+|v|<r\}.$ When $y\ne 0$  we have $r\leq |y| \implies B_d(\;(x,y),r)=\{(x,v): |v-y|<r.\quad$ And $r>|y|\implies B_d(\;(x.y),r)=\{(x,v) : |v-y|<r\}\cup B_d(\;(x,0),r-|y|).$
