Given:

The curve is x² + y² – 2x – 3 = 0

Differentiating the above w.r.t x, we get The Slope of tangent,

⇒ 2x^{2 – 1} + 2y^{2 – 1} – 2 – 0 = 0

⇒ 2x + 2y – 2 = 0

⇒ 2y = 2 – 2x

⇒ =

⇒ = ...(1)

(i) Since, the tangent is parallel to x – axis

⇒ = tan(0) = 0 ...(2)

tan(0) = 0

= The Slope of the tangent = tan

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

⇒ = 0

⇒ 1 – x = 0

⇒ x = 1

Substituting x = 1 in x² + y² – 2x – 3 = 0,

1² + y² – 2×1 – 3 = 0

1 + y² – 2 – 3 = 0

y² – 4 = 0

⇒ y² = 4

⇒ y = ±2

Thus, the required point is (1,2) & (1, – 2)