Find the points on the curve $x^2+y^2-2x-3=0$ at which the tangents are parallel to the X axis.
Differentiating the given curve w.r.t x gives
$2x+2y\dfrac{dy}{dx}-2=0$
i.e. $\dfrac{dy}{dx}=1-x$
Since the tangents are parallel to the X axis $\implies$ Slope of tangent $=0$
i.e. $\dfrac{dy}{dx}=0$
i.e. $1-x=0$ $\implies$ $x=1$
Substituting the value of x in the equation of curve gives
$1+y^2 -2 - 3 =0$
i.e. $y^2 = 4$ $\implies$ $y = \pm 2$
$\therefore$ $\;$ The required points are $\left(1,2\right)$ and $\left(1,-2\right)$