The equation of a common tangent to the parabolas $$y = x^2$$ and $$y = -(x-2)^2$$ is

We need to find the equation of a common tangent to the parabolas $$y = x^2$$ and $$y = -(x-2)^2$$.

At a point $$(t, t^2)$$ on $$y = x^2$$, the slope is $$\frac{dy}{dx} = 2t$$.

Tangent equation: $$y - t^2 = 2t(x - t)$$

$$y = 2tx - t^2$$ ... (i)

At a point $$(s, -(s-2)^2)$$ on $$y = -(x-2)^2$$, the slope is $$\frac{dy}{dx} = -2(s-2)$$.

Tangent equation: $$y + (s-2)^2 = -2(s-2)(x - s)$$

$$y = -2(s-2)x + 2s(s-2) - (s-2)^2$$

$$y = -2(s-2)x + (s-2)[2s - (s-2)]$$

$$y = -2(s-2)x + (s-2)(s+2)$$

$$y = -2(s-2)x + s^2 - 4$$ ... (ii)

Comparing slopes: $$2t = -2(s-2)$$

$$t = -(s-2) = 2 - s$$ ... (iii)

Comparing intercepts: $$-t^2 = s^2 - 4$$

$$t^2 + s^2 = 4$$ ... (iv)

Substitute $$t = 2 - s$$ into equation (iv):

$$(2-s)^2 + s^2 = 4$$

$$4 - 4s + s^2 + s^2 = 4$$

$$2s^2 - 4s = 0$$

$$2s(s - 2) = 0$$

$$s = 0 \text{ or } s = 2$$

When $$s = 0$$: $$t = 2$$, tangent: $$y = 4x - 4 = 4(x - 1)$$

When $$s = 2$$: $$t = 0$$, tangent: $$y = 0$$ (the x-axis)

$$y = 4(x-1)$$ matches Option B.

Therefore, the correct answer is Option B: $$y = 4(x-1)$$.