Question

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

• A. $(3,0),(-1,0)$
• B. $(3,0),(1,2)$
• C. $(-1,0),(1,2)$
• D. $(1,2),(1,-2)$

Answer 1

The correct option is D $(1,2),(1,-2)$

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

Differentiate w.r.t. x

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

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

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

$\frac{dy}{dx}=\frac{1-x}{y}$ ........$\left(1\right)$

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

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

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

$x=1$

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 $(1,-2)$.