Find the points on the curve

The equation of the given curve is $x^{2}+y^{2}-2 x-3=0$.

On differentiating with respect to x, we have:

$2 x+2 y \frac{d y}{d x}-2=0$

$\Rightarrow y \frac{d y}{d x}=1-x$

$\Rightarrow \frac{d y}{d x}=\frac{1-x}{y}$

Now, the tangents are parallel to the x-axis if the slope of the tangent is 0.

$\therefore \frac{1-x}{y}=0 \Rightarrow 1-x=0 \Rightarrow x=1$

But, $x^{2}+y^{2}-2 x-3=0$ for $x=1$

$\Rightarrow y^{2}=4 \Rightarrow y=\pm 2$

Hence, the points at which the tangents are parallel to the x-axis are (1, 2) and (1, −2).