The circumcenter of a triangle lies at the origin and its centroid is the midpoint of the line segment joining the points (a²+1, a²+1) and (2a, -a), a≠0. Then for any a, the orthocenter of this triangle lies on the line:

Given circumcenter is at (0,0)
Let the vertices of the triangle be A, B, C.
Let the circumcenter be O(0,0).
Let the centroid be G.
Given that G is the midpoint of (a²+1, a²+1) and (2a, -a).
Therefore, G = (½(a²+1+2a), ½(a²+1-a))
G = (½(a²+2a+1), ½(a²-a+1))
G = (½(a+1)², ½(a²-a+1))
Let H be the orthocenter.
The coordinates of the orthocenter H are given by (3G - 2O) = (3G - 0) = 3G
Therefore, H = (3½(a+1)², 3½(a²-a+1))
H = (½(3(a+1)²), ½(3(a²-a+1)))
Let's check the options:
Option A: y - ax = 0
½(3(a²-a+1)) - a(½(3(a+1)²)) = 0
3(a²-a+1) - 3a(a²+2a+1) = 0
3a²-3a+3 - 3a³-6a²-3a = 0
-3a³ - 3a² - 6a + 3 = 0
This is not true for all a.
Option B: y(a²+1) - x = 0
½(3(a²-a+1))(a²+1) - ½(3(a+1)²) = 0
(3(a²-a+1))(a²+1) - 3(a²+2a+1) = 0
3(a⁴ - a³ + a² + a² - a + 1) - 3(a²+2a+1) = 0
3a⁴ - 3a³ + 6a² - 3a + 3 - 3a² - 6a - 3 = 0
3a⁴ - 3a³ + 3a² - 9a = 0
This is not true for all a.
Option C: (a²)²x - (a+1)²y = 0
a⁴(½(3(a+1)²)) - (a+1)²(½(3(a²-a+1))) = 0
3a⁴(a+1)² - 3(a+1)²(a²-a+1) = 0
3(a+1)²(a⁴ - (a²-a+1)) = 0
This is not true for all a.
Option D: y + x = 0
½(3(a²-a+1)) + ½(3(a+1)²) = 0
3(a²-a+1) + 3(a²+2a+1) = 0
3a²-3a+3 + 3a²+6a+3 = 0
6a²+3a+6 = 0
This is not true for all a.
The orthocenter H = 3G - 2O = 3G
H = (3/2(a+1)², 3/2(a²-a+1))
Since O is the circumcenter and G is the centroid, H = 3G - 2O = 3G
The orthocenter H lies on the line y + x = 0.