At what points on the circle


At what points on the circle $x^{2}+y^{2}-2 x-4 y+1=0$, the tangent is parallel to $x$-axis?


Let the required point be (x1y1).
We know that the slope of the x-axis is 0.

$x^{2}+y^{2}-2 x-4 y+1=0$

$\left(x_{1}, y_{1}\right)$ lies on a curve.

$\therefore x_{1}^{2}+y_{1}^{2}-2 x_{1}-4 y_{1}+1=0$     ....(1)


$x^{2}+y^{2}-2 \mathrm{x}-4 y+1=0$

$\Rightarrow 2 x+2 y \frac{d y}{d x}-2-4 \frac{d y}{d x}=0$

$\Rightarrow \frac{d y}{d x}(2 y-4)=2-2 x$

$\Rightarrow \frac{d y}{d x}=\frac{2-2 x}{2 y-4}=\frac{1-x}{y-2}$

Slope of the tangent at $\left(x_{1}, y_{1}\right)=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{1-x_{1}}{y_{1}-2}$ $\ldots(2)$

Slope of the tangent $=0$ [Given]

$\therefore \frac{1-x_{1}}{y_{1}-2}=0$

$\Rightarrow 1-x_{1}=0$

$\Rightarrow x_{1}=1$

On substituting the value of $x_{1}$ in eq. (1), we get

$1+y_{1}^{2}-2-4 y_{1}+1=0$

$\Rightarrow y_{1}^{2}-4 y_{1}=0$

$\Rightarrow y_{1}\left(y_{1}-4\right)=0$

$\Rightarrow y_{1}=0,4$

Thus, the required points are $(1,0)$ and $(1,4)$.

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