1. **State the problem:** Find the points where the graph of the equation $$4x^2 + y^2 - 8x + 4y + 4 = 0$$ has a horizontal tangent line. 2. **Recall the formula:** A horizontal tangent line occurs where the derivative $$\frac{dy}{dx} = 0$$. 3. **Implicit differentiation:** Differentiate both sides with respect to $$x$$: $$\frac{d}{dx}(4x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(8x) + \frac{d}{dx}(4y) + \frac{d}{dx}(4) = 0$$ This gives: $$8x + 2y \frac{dy}{dx} - 8 + 4 \frac{dy}{dx} + 0 = 0$$ 4. **Group terms with $$\frac{dy}{dx}$$:** $$2y \frac{dy}{dx} + 4 \frac{dy}{dx} = 8 - 8x$$ 5. **Factor out $$\frac{dy}{dx}$$:** $$\frac{dy}{dx}(2y + 4) = 8 - 8x$$ 6. **Solve for $$\frac{dy}{dx}$$:** $$\frac{dy}{dx} = \frac{8 - 8x}{2y + 4}$$ 7. **Set $$\frac{dy}{dx} = 0$$ for horizontal tangent:** $$\frac{8 - 8x}{2y + 4} = 0 \implies 8 - 8x = 0 \implies x = 1$$ 8. **Find corresponding $$y$$ values:** Substitute $$x=1$$ into original equation: $$4(1)^2 + y^2 - 8(1) + 4y + 4 = 0$$ $$4 + y^2 - 8 + 4y + 4 = 0$$ $$y^2 + 4y = 0$$ 9. **Solve quadratic:** $$y(y + 4) = 0 \implies y = 0 \text{ or } y = -4$$ 10. **Final answer:** The points with horizontal tangent lines are $$\boxed{(1,0)}$$ and $$\boxed{(1,-4)}$$.