Given:

The curve is x² + y² – 2x – 4y + 1 = 0

Differentiating the above w.r.t x

⇒ x² + y² – 2x – 4y + 1 = 0

⇒ 2x^{2 – 1} + 2y^{2 – 1} – 2 – 4 + 0 = 0

⇒ 2x + 2y – 2 – 4 = 0

⇒ (2y – 4) = – 2x + 2

⇒

...(1)

= The Slope of the tangent = tan

Since, the tangent is parallel to x – axis

i.e,

⇒ = tan(0) = 0 ...(2)

tan(0) = 0

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

⇒ = 0

⇒ – (x – 1) = 0

⇒ x = 1

Substituting x = 1 in x² + y² – 2x – 4y + 1 = 0,we get,

⇒ 1² + y² – 21 – 4y + 1 = 0

⇒ 1 – y² – 2 – 4y + 1 = 0

⇒ y² – 4y = 0

⇒ y(y – 4) = 0

⇒ y = 0 & y = 4

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