Question

At what point, 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)

Given $x^2+y^2$ -2x -3 = 0 ......(1)
Diff. w.r.t.x, we get
2x + 2y$\frac{dy}{dx}$ - 2 = 0
$\Rightarrow \frac{dy}{dx}=\frac{1-x}{y}$
Since the slope of the tangent to the curve is zero.
$\therefore \frac{dy}{dx}$ = 0
$\Rightarrow$ 1 - x = 0
$\Rightarrow$ x = 1
Put x = 1 in equation (1), we get 1 + $y^2$ -2 -3 = 0
$\Rightarrow y^2$ = 4
$\Rightarrow$ y = $\pm 2$
Hence required points are (1, 2) and (1, -2)