How do uncoordinated turns (skid, slip) affect a helicopter's turn radius?
Given your assumptions about bank angle and airspeed and my assumption that coordinated turn is determined by acceleration inside aircraft pointing downward "perpendicular to the floor" (if you mean anything else by coordinated turn, try to be more specific on it), then you have one exact radius for coordinated turn. That one where centrifugal force generated by airspeed and curvature summed with gravity gives vector matching the (fixed) bank angle. If you want local acceleration to point in other direction (making slip or skid), only free variable remaining is turn radius.

So yes, given airspeed and bank angle is constant, turn radius is the parameter determining if turn is coordinated or not.

In a slip the local acceleration in aircraft points toward that side you want to turn (or, in other words, aircraft's floor is banked too much outside the turn). To get this, centrifugal force has to be less compared to coordinated turn (centrifugal force is only quality you allowed to change), so turn radius have to be larger. To get a skid turn radius have to be smaller for the same reason.

Inertial and gravity acceleration as seen and felt inside an aircraft during a turn with bank angle $\alpha$:

where $\vec{g}$ gravity acceleration and $\vec{a_c}$ centrifugal acceleration. Resulting $\vec{a}$ is total acceleration felt by pilot. From basic mechanics $a_c=v^2/r$ where $v$ and $r$ is speed (either TAS or GS, both are the same here) and turn radius respectively. Turn radius for coordinated turn is $\tan\alpha=a_c/g\Rightarrow r_{\rm coord}=v^2/(g\tan\alpha)$. Any smaller radius with the same bank and speed leads to higher $a_c$ and thus skid, larger radius results in slip. Note that this is true for any aircraft, not specific for helicopter.