Step-by-step Solution:

Let the common tangent be of the form \( y = mx + c \).\n\nFor the tangent to touch \( y = -x^2 \):\n\[ -x^2 = mx + c \implies x^2 + mx + c = 0 \]\nFor tangency, the discriminant must be zero:\n\[ m^2 - 4c = 0 \implies c = \frac{m^2}{4} \quad \text{--- (1)} \]\n\nFor the tangent to touch \( y = (x-2)^2 \):\n\[ (x-2)^2 = mx + c \implies x^2 - 4x + 4 = mx + c \]\n\[ x^2 - (m+4)x + (4-c) = 0 \]\nFor tangency, the discriminant must be zero:\n\[ (m+4)^2 - 4(4-c) = 0 \]\n\[ m^2 + 8m + 16 - 16 + 4c = 0 \]\n\[ m^2 + 8m + 4c = 0 \]\n\nSubstitute \( c = \frac{m^2}{4} \) from (1):\n\[ m^2 + 8m + 4\left(\frac{m^2}{4}\right) = 0 \]\n\[ 2m^2 + 8m = 0 \implies 2m(m+4) = 0 \]\n\nSo, \( m = 0 \) or \( m = -4 \).\nIf \( m = 0 \), \( c = 0 \). The tangent is \( y = 0 \).\nIf \( m = -4 \), \( c = \frac{(-4)^2}{4} = 4 \). The tangent is \( y = -4x + 4 \).\n\nThe correct option matching our result is \( y = -4x + 4 \).