What is $c$?

where is $c$?

Most probably it's just $y=ax^2+bx+c$ (there's no reason why the parabola shouldn't be opened down), but let 3333 confirm that.

It was actually $a+b+1=0$. Sorry for my mistake.

We have: $a+b+1=0 \iff 4a+4b+4=0 \iff a^2+4a+4b+4=a^2$ $ \iff (a+2)^2=a^2-4b \overset{a^2-4b>0}{\rightarrow} \iff 1=\frac{-a\pm\sqrt{a^2-4b}}{2}=x_{1,2}$ Otherwise, $B(\frac{-a+\sqrt{a^2-4b}}{2};0)$ and $C(\frac{-a-\sqrt{a^2-4b}}{2};0) $, where $B,C \equiv (P) \cap Ox$ and $A(0;b) \equiv (P) \cap Oy$ Let: $ I(m,m) \in d: y=x \implies d(I;AB)=d(I;AC)$, where $ I $ - incenter of $\bigtriangleup ABC$; $\implies f(\frac{-a\pm\sqrt{a^2-4b}}{2})=0 \implies f(1)=1+a+b=0$.

The statement is just wrong: two vertices of the triangle lie on the $x$ axis, thus the circle in question should be tangent to it. But then, if its center lies on $y=x$, it also must be tangent the to $y$ axis and the initial condition can't be met. The necessary amendment is to replace "incenter" by "circumcenter": The triangle vertices are $(x_1,0),(x_2,0),(0,b)$, hence if $O(m,m)$, then $m={x_1+x_2\over 2}=-{a\over 2}$ and $(m-x_1)^2+m^2=m^2+(m-b)^2$ $-2mx_1+x_1^2=-2mb+b^2$ $ax_1+x_1^2=ab+b^2\quad(*)$ By the initial condition, $x_1^2+ax_1=-b$, hence $(*)\implies b+ab+b^2=0\iff b(a+b+1)=0$ If $b=0$, then the triangle is degenerate, since two if its vertices coincide. Therefore $a+b+1=0$. QED

Incenter must be replaced by CIRCUMCENTER