The system x² + y² = 25 and y = x - 1 has two solutions. Which solution satisfies the additional constraint x > 0?

Substitute $y = x - 1$ into $x^2 + y^2 = 25$: $x^2 + (x-1)^2 = 25$, so $2x^2 - 2x + 1 = 25$ and $x^2 - x - 12 = 0$. This factors as $(x-4)(x+3) = 0$, so $x = 4$ or $x = -3$. The corresponding points are $(4, 3)$ and $(-3, -4)$, and only the first has $x > 0$. That is choice C. Choice A is the other valid intersection point, which has $x < 0$. Choices B and D are not solutions: $(3, 4)$ and $(-4, -3)$ both lie on the circle but have $y = x + 1$ instead of $y = x - 1$. Always verify a point in both equations.