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

Answer

The points where tangents are parallel to y-axis would be `=(-1,2),(3,2)`

Explanation

Given the Equation,

$$
\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)}
...