jmc

Let $r$ and $s$ be the $x$-coordinates of the two points of tangency. As such, they will be double roots of the polynomial $$(x^4+a{x}^3+x^2+bx+1)-x^2=x^4+a{x}^3+bx+1.$$Hence, $$\begin{array}{rl}{x}^4+a{x}^3+bx+1& =(x-r{)}^2(x-s{)}^2\\ & =(x^2-2rx+r^2)(x^2-2sx+s^2)\\ & =x^4-(2r+2s)x^3+(r^2+4rs+s^2)x^2-(2r^{2}s+2r{s}^2)x+r^2{s}^2.\end{array}$$Matching coefficients, we get $$\begin{array}{rl}{r}^2+4rs+s^2& =0,\\ r^2{s}^2& =1.\end{array}$$From $r^2{s}^2=1,$ either $rs=1$ or $rs=-1.$ But $4rs=-(r^2+s^2)$ is nonpositive, so $rs=-1.$ Then $$r^2+s^2=4.$$Hence, $(r-s{)}^2=r^2-2rs+s^2=6,$ so $|r-s|=\boxed{\sqrt{6}}.$