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 $(1,-2)$