At what points on the circle `x^2+y^2-2x-4y+1=0` , the tangent is parallel to the x-axis.

Answer

Points are `(1,0)` and `(1,4)`

Explanation

Let the point be `(x_{1}, y_{1})`.

We know that the slope of the `x`-axis is 0 .

So ATQ,


$$ x^{2}+y^{2}-2 x-4 y+1=0 $$
As `(x_{1}, y_{1})` lies on a curve.


`\therefore x_{1}^{2}+y_{1}^{2}-2 x_{1}-4 y_{1}+1=0 \ldots \text { (1) }`


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