Consider the locus equation  $[x^2+y^2]-x-y=0$ ([.] denotes G.I.F) which consists of line segments when plotted in Cartesian plane. If the sum of lengths of all possible line segment is given by  $\sqrt{a}+\sqrt{b}-\sqrt{c}(a,b,c\in N).$ Then  $a+b+c=$

We split into cases on the integer  $k=[x^2+y^2].$ Note that  $x+y=k$  but  $x^2+y^2\ge \frac{1}{2}{(x+y)}^2=\frac{1}{2}{k}^2$  and  $x^2+y^2<1,$  which forces $k\le 2.$ 
If  $k=0,$  the region defined by  $0\le x^2+y^2<1$ and $x+y=0$ is the diameter from  $(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$  to  $(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}),$ which has length 2.
If  $k=1,$  the region  $1\le x^2+y^2<2$ and  $x+y=1$ consists of two segments, which is the chord on  $x^2+y^2=2$ minus the chord on  $x^2+y^2=1.$ The former has length  $2\sqrt{{\left(\sqrt{2}\right)}^2-{\left(\frac{\sqrt{2}}{2}\right)}^2}=\sqrt{6},$  and the latter has length  $2\sqrt{1^2-{\left(\frac{\sqrt{2}}{2}\right)}^2}=\sqrt{2}.$  So the total length here is  $\sqrt{6}-\sqrt{2}.$
If $k=2,$  the region  $2\le x^2+y^2<3$  and  $x+y=1$  is the chord on $x^2+y^2=3,$  which has length  $2\sqrt{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}=2.$
Our final answer is $2+(\sqrt{6}-\sqrt{2})+2=4+\sqrt{6}-\sqrt{2}.$