At what points on the curve `x^2+y^2-2x-4y+1=0`, the tangents are parallel to the `y-a` is?

Answer

Points are `= (-1,2),(3,2)`

Explanation

$$ \begin{aligned} &x^{2}+y^{2}-2 x-4 y+1 \\ &\Rightarrow 2 x+2 y \frac{d y}{d x}-2-4 \frac{d y}{d x}=0 \\ &\Rightarrow x+y \frac{d y}{d x}-1-2 \frac{d y}{d x}=0 \Rightarrow(y-2) \frac{d y}{d x}=(1-x) \\ &\frac{d y}{d x}=\frac{(y-2)}{(1-x)} \end{aligned} $$
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