Consider the curve given by the equation $x^2 + y^2 = 2x + 4y - 4$. At what points on the curve is the tangent line horizontal?

Core Concept

Correct. By completing the square, the equation can be rewritten as $(x-1)^2 + (y-2)^2 = 1$, which is a circle centered at $(1, 2)$ with a radius of 1. The horizontal tangent lines of a circle occur at the points directly above and below the center, which are $(1, 1)$ and $(1, 3)$. Equivalently, using implicit differentiation yields $y' = (1-x)/(y-2)$, which is zero when $x = 1$; substituting $x = 1$ into the original equation yields $y = 1$ and $y = 3$.