Question

At what points on the curve $x^2+y^2-2x-4y+1=0$, the tangents are parallel to the $Y-axis$?

Answer 1

Given equation,
$x^2+y^2-2x-4y+1=0$.................(1)
can be written as,
$x^2-2x+1+y^2-4y+4=4$
$or,(x-1{)}^2+(y-2{)}^2=2^2$
Which represents a circle of radius $2$ and centre at $(1,2)$.
As the points where the tangent is parallel to the $Y-axis$ for the circle $x^2+y^2=2^2$................(2)
are $(-2,0)$ and $(2,0)$.
So, the points where the tangent is parallel to the $Y-axis$ for the circle $(x-1{)}^2+(y-2{)}^2=2^2$ will be $(-2+1,0+2)$ and $(2+1,0+2)$ i.e. $(-1,2)$ and $(3,2)$.
(When the origin is shifted to $(1,2)$ for the circle (2) then we get circle (1)).