(c) $(-2, 0)$
The centre $(2, 0)$ is the midpoint of the diameter. If one end is $(6, 0)$ and the other end is $(x, y)$, then $\frac{6+x}{2} = 2 \Rightarrow x = -2$ and $y = 0$.
Use the midpoint formula: centre = midpoint of diameter. Set $\frac{x_1+x_2}{2} = 2$ with $x_1 = 6$, giving $x_2 = -2$. This is a standard 1-mark application of the midpoint concept from Chapter 7.