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:

\[x^2 + y^2 - 2x - 4y + 1 = 0 \]

\[\left( x_1 , y_1 \right) \text { lies on a curve .} \]

\[ \therefore {x_1}^2 + {y_1}^2 - 2 x_1 - 4 y_1 + 1 = 0 . . . \left( 1 \right)\]

\[\text { Now,} \]

\[ x^2 + y^2 - 2x - 4y + 1 = 0 \]

\[ \Rightarrow 2x + 2y \frac{dy}{dx} - 2 - 4\frac{dy}{dx} = 0\]

\[ \Rightarrow \frac{dy}{dx} \left( 2y - 4 \right) = 2 - 2x\]

\[ \Rightarrow \frac{dy}{dx} = \frac{2 - 2x}{2y - 4} = \frac{1 - x}{y - 2}\]

\[\text { Slope of the tangent at }\left( x_1 , y_1 \right)= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =\frac{1 - x_1}{y_1 - 2}...(2)\]

\[\text { Slope of the tangent } = 0 [\text { Given }]\]

\[ \therefore \frac{1 - x_1}{y_1 - 2} = 0\]

\[ \Rightarrow 1 - x_1 = 0\]

\[ \Rightarrow x_1 = 1\]

\[\text { On substituting the value of } x_1 \text { in eq. (1), we get }\]

\[1 + {y_1}^2 - 2 - 4 y_1 + 1 = 0\]

\[ \Rightarrow {y_1}^2 - 4 y_1 = 0\]

\[ \Rightarrow y_1 \left( y_1 - 4 \right) = 0\]

\[ \Rightarrow y_1 = 0, 4\]

\[\text { Thus, the required points are (1, 0) and (1, 4)}.\]