1. **Problem statement:** Find the slope $m$ of the line $y=mx$ that is tangent to the circle given by $$(x-2)^2 + (y-6)^2 = 4.$$\n\n2. **Formula and concept:** A line is tangent to a circle if it touches the circle at exactly one point. This means the system of equations formed by the line and the circle has exactly one solution.\n\n3. **Substitute $y=mx$ into the circle equation:**\n$$ (x-2)^2 + (mx - 6)^2 = 4 $$\nExpanding:\n$$ (x-2)^2 + (mx - 6)^2 = (x-2)^2 + (m^2x^2 - 12mx + 36) = 4 $$\n\n4. **Expand and simplify:**\n$$ (x^2 - 4x + 4) + (m^2x^2 - 12mx + 36) = 4 $$\n$$ x^2 - 4x + 4 + m^2x^2 - 12mx + 36 = 4 $$\n$$ (1 + m^2)x^2 - (4 + 12m)x + (4 + 36 - 4) = 0 $$\n$$ (1 + m^2)x^2 - (4 + 12m)x + 36 = 0 $$\n\n5. **Condition for tangency:** The quadratic in $x$ must have exactly one solution, so its discriminant $D$ must be zero:\n$$ D = b^2 - 4ac = 0 $$\nHere, $a = 1 + m^2$, $b = -(4 + 12m)$, $c = 36$.\n\n6. **Calculate discriminant:**\n$$ D = (-(4 + 12m))^2 - 4(1 + m^2)(36) = (4 + 12m)^2 - 144(1 + m^2) $$\n\n7. **Expand and simplify:**\n$$ (4 + 12m)^2 - 144 - 144m^2 = 0 $$\n$$ 16 + 96m + 144m^2 - 144 - 144m^2 = 0 $$\n$$ 16 + 96m - 144 = 0 $$\n$$ 96m - 128 = 0 $$\n\n8. **Solve for $m$:**\n$$ 96m = 128 $$\n$$ m = \frac{128}{96} = \frac{4}{3} $$\n\n**Final answer:** $m = \frac{4}{3}$, which corresponds to option (d).