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# At what point on the circle

Question:

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

Solution:

Given:

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

Differentiating the above w.r.t $x$

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

$\Rightarrow 2 x^{2}-1+2 y^{2}-1 \times \frac{d y}{d x}-2-4 x \frac{d y}{d x}+0=0$

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2(\mathrm{x}-1)}{2(\mathrm{y}-2)}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-(\mathrm{x}-1)}{(\mathrm{y}-2)} \ldots(1)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=$ The Slope of the tangent $=\tan \theta$

Since, the tangent is parallel to $x$ - axis

i.e,

$\Rightarrow \frac{d y}{d x}=\tan (0)=0 \ldots(2)$

$\therefore \tan (0)=0$

From (1) & (2), we get,

$\Rightarrow \frac{-(x-1)}{(y-2)}=0$

$\Rightarrow-(x-1)=0$

$\Rightarrow x=1$

Substituting $x=1$ in $x^{2}+y^{2}-2 x-4 y+1=0$, we get,

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

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

$\Rightarrow y^{2}-4 y=0$

$\Rightarrow y(y-4)=0$

$\Rightarrow y=0 \& y=4$

Thus, the required point is $(1,0) \&(1,4)$