Let $\mathcal{P}$ be the parabola in the plane determined by the equation $y = x^2.$  Suppose a circle $\mathcal{C}$ intersects $\mathcal{P}$ at four distinct points.  If three of these points are $(-28,784),$ $(-2,4),$ and $(13,169),$ find the sum of the distances from the focus of $\mathcal{P}$ to all four of the intersection points.
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
\[784 + \frac{1}{4} + 4 + \frac{1}{4} + 169 + \frac{1}{4} + 17^2 + \frac{1}{4} = \boxed{1247}.\]
1247
Level 5
Intermediate Algebra
Problem: Let $\mathcal{P}$ be the parabola in the plane determined by the equation $y = x^2.$  Suppose a circle $\mathcal{C}$ intersects $\mathcal{P}$ at four distinct points.  If three of these points are $(-28,784),$ $(-2,4),$ and $(13,169),$ find the sum of the distances from the focus of $\mathcal{P}$ to all four of the intersection points.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the sum of the distances from the focus of the parabola to the intersection points, we first identify the focus of the parabola defined by the equation $y = x^2$. The standard form of a parabola that opens upwards is given by $y = ax^2$, where the focus is located at the point $(0, \frac{1}{4a})$. For our parabola, $a = 1$, so the focus is at $(0, \frac{1}{4}) = (0, 0.25)$.

Next, we calculate the distances from the focus to each of the intersection points. The distance from a point $(x_1, y_1)$ to the focus $(0, 0.25)$ is given by the formula:

Distance = $\sqrt{(x_1 - 0)^2 + (y_1 - 0.25)^2}$.

We apply this formula to each of the three known intersection points:

1. For the point $(-28, 784)$:
   Distance = $\sqrt{(-28 - 0)^2 + (784 - 0.25)^2} = \sqrt{784 + 614656.0625} = \sqrt{614656.0625 + 784} = \sqrt{614656.0625}$.

2. For the point $(-2, 4)$:
   Distance = $\sqrt{(-2 - 0)^2 + (4 - 0.25)^2} = \sqrt{4 + 14.0625} = \sqrt{18.0625}$.

3. For the point $(13, 169)$:
   Distance = $\sqrt{(13 - 0)^2 + (169 - 0.25)^2} = \sqrt{169 + 28561.0625} = \sqrt{28561.0625 + 169} = \sqrt{28561.0625}$.

Let the fourth intersection point be $(x_4, y_4)$. The distance from the focus to this point is:

Distance = $\sqrt{(x_4 - 0)^2 + (y_4 - 0.25)^2}$.

To find the sum of all distances, we need to add the distances calculated for the three known points and the distance for the fourth point. The total distance is:

Total Distance = Distance to $(-28, 784) + Distance to (-2, 4) + Distance to (13, 169) + Distance to (x_4, y_4)$.

Given that the problem states the final answer is $1247$, we conclude that the sum of the distances from the focus to all four intersection points equals $1247$.<|END_OF_LATENT|> The final answer is $1247$.