The curve C has equation y²-4y-4x-12. (a) Find the coordinates of point A at which C meets the x-axis. (b) Show that the equation can be written in the form (y−2)²=4(x−2). (c) Show that a point Q with coordinates (2+r²;2+2r) lies on C. (d) Find the equation of the tangent at A.

The coordinates of point A are $\boxed{(-3, 0)}$.

Add $\left(\frac{-4}{2}\right)^2 = (-2)^2 = 4$ to both sides.

This result, $(y-2)^2 = 4(x+4)$, is not in the form $(y-2)^2 = 4(x-2)$. There is an inconsistency in the question. The given equation $y^2 - 4y - 4x - 12 = 0$ does not transform into $(y-2)^2 = 4(x-2)$. It transforms into $(y-2)^2 = 4(x+4)$. For the equation to be $(y-2)^2 = 4(x-2)$, the original equation should have been $y^2 - 4y - 4x + 12 = 0$.
For parts (c) and (d), we will assume the intended equation of the curve is $(y-2)^2 = 4(x-2)$ as implied by the "show that" part.

Since LHS = RHS ($4r^2 = 4r^2$), the point Q with coordinates $(2+r^2, 2+2r)$ lies on the curve C.

Set $y=0$: $(0-2)^2 = 4(x-2) \implies 4 = 4(x-2) \implies 1 = x-2 \implies x=3$.
So, point A is $(3,0)$.

The equation is $(y-2)^2 = 4(x-2)$.
Differentiating both sides:

Substitute $y=0$ into the derivative:

The gradient of the tangent at A is $-1$.

Using point A $(3,0)$ and gradient $m=-1$:

The equation of the tangent at A is $\boxed{x + y = 3}$.