we need to find a mapping of those points onto $\mathbf{R}^2$ so that the two colors classes are can be separated by a line?
So I don't know the radius of those circles. But if I knew, I would take their norm. However that would be a mapping to $\mathbf{R}$.
On
I'm not entirely clear on the constraints of the problem, but the following may be suggestive:
Assume you have rectangular-to-polar conversions $r(x,y)$ and $\theta(x,y)$. Let $P$ be the coordinates of the positive class and $N$ be the coordinates of the negative class. Then in your terms, $r_2=\min_{(x,y)\in P} \left\{r(x,y)\right\}$ (the positive examples all have the same radius). A possible mapping is $(x,y)\mapsto(\left|r(x,y)-r_2\right|,\theta(x,y))$, in which case a valid linear separator is the vertical line at $\frac{1}{2}\min_{(x,y)\in N} \left\{\left|r(x,y)-r_2\right|\right\}$.
If we don't require the map to be one to one, then $f((x,y))=(1,0)$ for points on the outer and inner circle and $f((x,y))=(0,1)$ for points on the middle circle would work. If we need the map to be one to one, then we can map the circles to an interval like $(0,2\pi)$ would be something to try. Let $r_1,r_2,r_3$ be the radius of inner, middle and outer circle respectively. Then points on the circle are represented using by the function $r_me^{i\theta}$.So our $2D$ linearly separable function could be that we map the points on the inner circle $r_1e^{i\theta_1}$to $(1,\theta_1)$ and the points on the outer circle to $r_3e^{i\theta_3}$ to $(2,\theta_3)$ and points on the middle circles $r_2e^{i\theta_2}$ to $(3,\theta_2)$. The rest of the point on the plane can be mapped to say $(100,100)$ and we get a linearly separable 2D mapping!