Given:

Curves x² + y² – 4x – 1 = 0 ...(1)

& x² + y² – 2y – 9 = 0 ...(2)

First curve is x² + y² – 4x – 1 = 0

⇒ x² – 4x + 4 + y² – 4 – 1 = 0

⇒ (x – 2)² + y² – 5 = 0

Now ,Subtracting (2) from (1),we get

⇒ x² + y² – 4x – 1 – ( x² + y² – 2y – 9) = 0

⇒ x² + y² – 4x – 1 – x² – y² + 2y + 9 = 0

⇒ – 4x – 1 + 2y + 9 = 0

⇒ – 4x + 2y + 8 = 0

⇒ 2y = 4x – 8

⇒ y = 2x – 4

Substituting y = 2x – 4 in (3),we get,

⇒ (x – 2)² + (2x – 4)² – 5 = 0

⇒ (x – 2)² + 4(x – 2)² – 5 = 0

⇒ (x – 2)²(1 + 4) – 5 = 0

⇒ 5(x – 2)² – 5 = 0

⇒ (x – 2)² – 1 = 0

⇒ (x – 2)² = 1

⇒ (x – 2) = ±1

⇒ x = 1 + 2 or x = – 1 + 2

⇒ x = 3 or x = 1

So ,when x = 3

y = 2×3 – 4

⇒ y = 6 – 4 = 2

So ,when x = 3

y = 2×1 – 4

⇒ y = 2 – 4 = – 2

The point of intersection of two curves are (3,2) & (1, – 2)

Now ,Differentiating curves (1) & (2) w.r.t x, we get

⇒ x² + y² – 4x – 1 = 0

⇒ 2x + 2y.4 – 0 = 0

⇒ x + y.2 = 0

⇒ y. = 2 – x

...(3)

⇒ x² + y² – 2y – 9 = 0

⇒ 2x + 2y.20 = 0

⇒ x + y.0

⇒ x + (y – 1)0

⇒ ...(4)

At (3,2) in equation(3),we get

m₁

At (3,2) in equation(4),we get

= – 3

m₂ = – 3

when m₁ & m₂ = 0

tanθ

tanθ

tanθ

tanθ = 7

θ = tan ^{– 1}(7)

θ≅81.86