At what points on the curve $x^{2}+y^{2}-2x-4y+1=0$,are the tangents parallel to the $y$-axis?

(A) Given the equation of the curve: $x^{2}+y^{2}-2x-4y+1=0$ ... $(i)$
Differentiating with respect to $x$:
$2x + 2y \frac{dy}{dx} - 2 - 4 \frac{dy}{dx} = 0$
$\frac{dy}{dx}(2y - 4) = 2 - 2x$
$\frac{dy}{dx} = \frac{2(1-x)}{2(y-2)} = \frac{1-x}{y-2}$
Since the tangents are parallel to the $y$-axis,the slope $\frac{dy}{dx}$ is undefined,which means the denominator must be zero:
$y - 2 = 0 \Rightarrow y = 2$
Substituting $y = 2$ into the original equation $(i)$:
$x^{2} + (2)^{2} - 2x - 4(2) + 1 = 0$
$x^{2} + 4 - 2x - 8 + 1 = 0$
$x^{2} - 2x - 3 = 0$
Factoring the quadratic equation:
$(x - 3)(x + 1) = 0$
$x = 3$ or $x = -1$
Thus,the required points are $(3, 2)$ and $(-1, 2)$.