Consider the locus equation  $[x^2+y^2]-x-y=0$  $\left([.]denotesG.I.F\right)$  which consists of line segments when plotted in Cartesian plane. If the sum of lengths of all possible line segment is given by  $\sqrt{a}+\sqrt{b}-\sqrt{c}(a,b,c\in N)$  find 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  $2\sqrt{{\left(\sqrt{2}\right)}^2-{\left(\frac{\sqrt{2}}{2}\right)}^2}=\sqrt{6},$ and the latter has length  $2\sqrt{{\left(\sqrt{3}\right)}^2-{\left(\frac{\sqrt{3}}{2}\right)}^2}=\sqrt{2}.$ 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 $2+(\sqrt{6}-\sqrt{2})+2=4+\sqrt{6}-\sqrt{2}$