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**redsoxfan325** Note that $\displaystyle x^2+y^2-6x+10y=20 \implies x^2-6x+9-9+y^2+10y+25-25=20$ $\displaystyle \implies (x-3)^2+(y+5)^2=54$

So the function you are dealing with is a circle of radius $\displaystyle 3\sqrt{6}$ centered at $\displaystyle (3,-5)$. This may be a helpful picture to have in your mind while doing this.

Given that $\displaystyle \frac{dy}{dx}=\frac{6-2x}{2y+10}$, the horizontal tangent occurs when the *numerator* is zero (because then the derivative is 0), which is at $\displaystyle x=3$. Use the function to find that $\displaystyle y=-5\pm 3\sqrt{6}$.

Similarly, the vertical tangent occurs when the *denominator* is zero (because then the derivative is "infinity"), which is at $\displaystyle y=-5$. Use the function to find that $\displaystyle x=3\pm 3\sqrt{6}$.

So the horizontal tangent lines occur at $\displaystyle (3,-5+3\sqrt{6})$ and $\displaystyle (3,-5-3\sqrt{6})$ and the vertical tangent lines occur at $\displaystyle (3+3\sqrt{6},-5)$ and $\displaystyle (3-3\sqrt{6},-5)$.

If you think about the picture of the circle, this makes a lot of sense.