jmc

Let the four intersection points be $(a,a^2),$ $(b,b^2),$ $(c,c^2),$ and $(d,d^2).$ Let the equation of the circle be $$(x-k{)}^2+(y-h{)}^2=r^2.$$Substituting $y=x^2,$ we get $$(x-k{)}^2+(x^2-h{)}^2=r^2.$$Expanding this equation, we get a fourth degree polynomial whose roots are $a,$ $b,$ $c,$ and $d.$ Furthermore, the coefficient of $x^3$ is 0, so by Vieta's formulas, $a+b+c+d=0.$

We are given that three intersection points are $(-28,784),$ $(-2,4),$ and $(13,196),$ so the fourth root is $-((-28)+(-2)+13)=17.$

The distance from the focus to a point on the parabola is equal to the distance from the point to the directrix, which is $y=-\frac{1}{4}.$ Thus, the sum of the distances is $$784+\frac{1}{4}+4+\frac{1}{4}+169+\frac{1}{4}+17^2+\frac{1}{4}=\boxed{1247}.$$