# At what points the slope of the tangent to the curve

Question:

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

A. $(3,0),(-1,0)$

B. $(3,0),(1,2)$

C. $(-1,0),(1,2)$

D. $(1,2),(1,-2)$

Solution:

Given that the curve $x^{2}+y^{2}-2 x-3=0$

Differentiation on both the sides,

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1-\mathrm{x}}{\mathrm{y}}$

According to the question,

Slope of the tangent $=0$

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

$\Rightarrow x=1$

Putting this in equation of curve,

$1+y^{2}-2-3=0$

$\Rightarrow y^{2}=4$

$\Rightarrow y=\pm 2$

So, the required points are $(1,2)$ and $(1,-2)$