Question

At what point the slope of the tangent to the curve x² + y² − 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

(d) (1, 2), (1, −2)

Let (x₁, y₁) be the required point.

$$\begin{gathered} \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lie}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},{x_1}^2+{y_1}^2-2x_1-3=0...\left(1\right) \\ \mathrm{Now},x^2+y^2-2x-3=0 \\ \Rightarrow 2x+2y\frac{dy}{dx}-2=0 \\ \therefore \frac{dy}{dx}=\frac{2-2x}{2y}=\frac{1-x}{y} \\ \mathrm{Now}, \\ \text{Slope of the tangent =}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=}\frac{1-x_1}{y_1} \\ \mathrm{S}\text{lope of the tangent=0 (Given)} \\ \therefore \frac{1-x_1}{y_1}=0 \\ \Rightarrow 1-x_1=0 \\ \Rightarrow x_1=1 \\ \text{From (1), we get} \\ {x_1}^2+{y_1}^2-2x_1-3=0 \\ \Rightarrow 1+{y_1}^2-2-3=0 \\ \Rightarrow {y_1}^2-4=0 \\ \Rightarrow y_1=\pm 2 \\ \text{So, the points are}\left(1,2\right)\text{and}\left(1,-2\right)\text{.} \end{gathered}$$