At what points on the curve `x^2+y^2-2x-4y+1=0` , the tangents are parallel to the `y-a xi s ?`

Explanation

Given, equation of curve which is
`x^(2) + y^(2)-2x-4y+1=0`
`rArr 2x+2y(dy)/(dx)-2-4(dy)/(dx)=0`
`rArr (dy)/(dx)(2y-4)=2-2x`
`rArr (dy)/(dx) = (2(1-x))/(2(y-2))`
Since, the tangents are parall to the Y-axis i.e., `tantheta = tan90^(@)=(dy)/(dx)`
`therefore (1-x)/(y-2)=1/0`
`rArr y-2=0`
`rArr y=2`
For y=2 from Eq.(i), we get
`x^(2)+2^(2)-2x-4 xx 2+1=0`
`rArr x^(2)-2x-3=0`
`rArr x(x-3)+1(x-3)=0`
`rArr (x+1)(x-3)=0`
`therefore x=-1, x=3`
So, the required points are `(-1,2)` and (3,2).