The equation of the tangent to the parabola \( y = x^2 \) at \( (t, t^2) \) is \( tx = y + at^2 \). Substituting \( y = x^2 \), we get \( y = tx - \frac{t^2}{4} \). Equating this with \( y = (x - 2)^2 \) for a common tangent yields \( tx - \frac{t^2}{4} = (x - 2)^2 \). Expanding and simplifying leads to \( x^2 + x(t - 4) - \frac{t^2}{4} + 4 = 0 \). For a common tangent, the discriminant must be zero: \( (t - 4)^2 - 4 \left( 4 - \frac{t^2}{4} \right) = 0 \). Solving for \( t \) gives \( t^2 - 4t = 0 \), so \( t = 0 \) or \( t = 4 \). Substituting \( t = 4 \) into the tangent equation results in \( y = 4(x - 1) \). The correct answer is \(\boxed{(b)}\).