jmc

$y=7$ represents a horizontal line that intersects a circle of radius $10$ around the origin. The symmetry of the circle guarantees that the points of intersection have a sum which add up to $0.$

Alternatively, we can simply substitute $7$ into the second equation for $y,$ to get that $x^2=51.$ Then, the two possible values for $x$ are $\sqrt{51},-\sqrt{51}.$ It's clear that they add up to $\boxed{0}.$