Is there any Group structure on $\Bbb R^2$ with Disks subgroups? and ... But is there any group structure on $\Bbb R^2$ such that its proper subgroups are only Disks with center $(0,0)$? (subgroups are $\{(x.y)\,|\, x^2+y^2 \le r^2 \}$ and $(0,0)$ is identity element!)
Can a bounded subset of $\Bbb R^2$ be a Group? I believe each set can't be a Group like $S^2$ (maybe this is a discussion in Universal algebra further!)
I thank you in advance.
 A: 
But is there any group structure on $\Bbb R^2$ such that its real subgroups are only circles with center $(0,0)$?

There are a couple barriers that come to mind.
First, the cyclic group generated by a non-identity element will be a proper subgroup that is at most countable. But circles in the plane are uncountable.
Second, all subgroups must contain the identity, so all subgroups must intersect in at least one point. But two different concentric circles do not intersect.

Can a bounded subset of $\Bbb R^2$ be a Group?

Sure, if we're choosing an arbitrary group structure! Just place $\mathbb R^2$ in bijection with the Cantor group $2^\omega$, which has tons of finite subgroups.
A: If you want all subgroups to be circles(or disks or any uncountable sets) centered at zero then such a group does not exists. This is because all uncountable groups contain countable subgroups, for example another answer here points out such groups have cyclic subgroups.
A nice example is choose a countable subset, like $\mathbb{Q}^2$, and note the group generated by $\mathbb{Q}^2$, $H= \langle \mathbb{Q}^2 \rangle$, will be a countable subgroup of $(\mathbb{R}^2, \star)$, where "$\star$" is any group operation on the set $\mathbb{R}^2$. This subgroup has the property that it is not contained in any bounded set.
Note that you can have subgroups which are bounded. Another answer points out you can put a group structure on $\mathbb{R}^2$ so that it has many finite subgroups, which have to be bounded. You can have bounded infinite groups though. As an example consider $\mathbb{R}^2$ under addition and the subgroups which are lines through the origin. You can put these in bijection with the circles plus origin, and transfer the group structure over, so now circles with origin are bounded subgroups isomorphic to the real line under addition.
Note the above does not take any topological considerations into account.
