Question

At what points on the circle x² + y² − 2x − 4y + 1 = 0, the tangent is parallel to x-axis?

Answer 1

Let the required point be (x₁, y₁).
We know that the slope of the x-axis is 0.
Given:

$$\begin{gathered} x^2+y^2-2x-4y+1=0 \\ \left(x_1,y_1\right)\mathrm{lies}\mathrm{on}\mathrm{a}\mathrm{curve}. \\ \therefore {x_1}^2+{y_1}^2-2x_1-4y_1+1=0...\left(1\right) \\ \mathrm{Now}, \\ {x}^2+y^2-2\mathrm{x}-4y+1=0 \\ \Rightarrow 2x+2y\frac{dy}{dx}-2-4\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}\left(2y-4\right)=2-2x \\ \Rightarrow \frac{dy}{dx}=\frac{2-2x}{2y-4}=\frac{1-x}{y-2} \\ \text{Slope of the tangent at}\left(x_1,y_1\right)\text{=}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=}\frac{1-x_1}{y_1-2}\text{...(2)} \\ \text{Slope of the tangent = 0 [Given]} \\ \therefore \frac{1-x_1}{y_1-2}=0 \\ \Rightarrow 1-x_1=0 \\ \Rightarrow x_1=1 \\ \text{On substituting the value of}{x}_1\text{in eq. (1), we get} \\ 1+{y_1}^2-2-4y_1+1=0 \\ \Rightarrow {y_1}^2-4y_1=0 \\ \Rightarrow y_1\left(y_1-4\right)=0 \\ \Rightarrow y_1=0,4 \\ \text{Thus, the required points are (1, 0) and (1, 4).} \end{gathered}$$