At what points the slope of the tangent to the curve $x^2+y^2-2x-3=0$ is zero?

$x^2+y^2-2x-3=0$ is zero.

Differentiate w.r.t. x

$2x+2y\dfrac{dy}{dx}-2=0$

$2y\cdot \dfrac{dy}{dx}=2-2x$

$\dfrac{dy}{dx}=\dfrac{2(1-x)}{2y}$

$\dfrac{dy}{dx}=\dfrac{1-x}{y}$ ........$(1)$

If line is parallel to x-axis

Angle with x-axis $=\theta =0$

Slope of x-axis$=\tan\theta =\tan 0^o=0$

Slope of tangent $=$ Slope of x-axis

$\dfrac{dy}{dx}=0$

$\dfrac{1-x}{y}=0$

$x=1$

Finding y when $x=1$

$x^2+y^2-2x-3=0\\$

$(1)^2+y^2-2(1)-3=0\\$

$1+y^2-2-3=0\\$

$y=\pm 2$

Hence, the points are $(1, 2)$ and $(1, -2)$.