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Use implicit differentiation to find the points where the circle defined by x2+y2−2x−4y=−1 has horizontal tangent lines. List your answers as points in the form (a,b). 1. Find the points where the curve has a horizontal tangent line.

How do I solve this question? I know i have to solve for y' to find the gradient of the slope which I calculate to be y' = (2x-2)/(2y-4) y' = x-1/y-2

What do I do after this?

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First note that $dy/dx = \color{red}{-}(x-1)/(y-2)$. Since you're looking for where the circle has horizontal tangents, set the numerator of $dy/dx$ equal to zero and solve for $x$. Then plug this value back into your implicitly defined function (the circle) and solve for $y$. You should get two different $y$ values. – Christopher Toni Oct 24 '13 at 10:31

Your circle $\gamma$ is a level line of the function $$F(x,y):=x^2+y^2-2x-4y\ .$$ Therefore at any point $P=(x,y)\in\gamma$ the gradient $\nabla F(P)=(2x-2,2y-4)$ is orthogonal to the tangent at $P$. Since we want the points $P\in\gamma$ where the tangent is horizontal we want the points where the normal $\nabla F(P)$ is vertical, i.e., where $2x-2=0$, or $x=1$. Putting $x=1$ in the equation for $\gamma$ leads to the equation $$1+y^2-2-4y=-1$$ with the two solutions $y_1=0$ and $y_2=4$. It follows that there are two points where $\gamma$ has a horizontal tangent, namely the points $P_1=(1,0)$ and $P_2=(1,4)$.

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