Application of Derivatives — Class 12 Maths Solution

${x^2} + {y^2} - 2x - 3 = 0$ …(i)

Differentiating (i) w. r.t. $x$, we get
$2x + 2y\cfrac{{dy}}{{dx}} - 2 - 0 = 0 \Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{{2(1 - x)}}{{2y}} = \cfrac{{1 - x}}{y}$ …(ii)

For tangents parallel to x-axis, we must have
$\cfrac{{dy}}{{dx}} = 0 \Rightarrow \cfrac{{1 - x}}{y} = 0\; \Rightarrow x = 1,y \ne 0$

Substituting $x = 1$ in (i), we get
${1^2} + {y^2} - 2 \cdot 1 - 3 = 0 \Rightarrow {y^2} - 4 = 0 \Rightarrow y = \pm \,\,2$

Hence, the required points are $\left( {1,2} \right)$ and $\left( {1, - 2} \right)$.