Summary:
There are global parameters, $R$ - thickness of red area, $G$ - thickness of green area, $S=\frac{R}{R+G}$ - represents start signed distance of red color, $E=\frac{0.5}{R+G}$ - represents end signed distance of red color. Each point also has own parameter $F_{(x,y)}$ - represents 2D field of signed distance, and it is triangulary interpolated by: $L\triangle{IBO}$ $L\triangle{IOR}$ $R\triangle{ILO}$ $R\triangle{IOB}$ and $\triangle{I_{L}R_{L}L_{R}}$ $\triangle{I_{L}L_{R}I_{R}}$ triangles. The values of it are:
$F_{LR}={(1, 0)}; F_{LI}={(0, 0)}; F_{LB}={(1, 0)}; F_{LO}={(1, 1)};$
$F_{RL}={(1, 0)}; F_{RI}={(1-S, 0)}; F_{RB}={(1, 0)}; F_{RO}=?;$
Distance at each pixel is calculated by: $d=1-\sqrt{{F_{x}}^2+{F_{y}}^2}$.
If $d_x < S$ then color of pixel is green, else if $d_x \geq S; d_x \leq E$ color of pixel is red.
Issue: So, judging by the picture, it's clear that i'm trying to get rid of the green area on the right side, and it's done by changing $F_{RI}$ but $F_{RO}=(1, 1)$ makes arc looking wrong by going out of bounds (at the picture i shown expected result, not the current). I guess what is needed to find is $F_{RO}$ that will make same arc as on the left side (of course with a less radius, due to size difference) and not being breaked by interpolation of parameters.