Question

If two circles `x^2 + y^2 + 2n_1x + 2y + 1/2` = 0 and `x^2 + y^2 + n_2x + n_2y + n_1 = 1/2`, intersect each other orthogonally where n₁, n₂ ∈ I, then number of possible of ordered pairs (n₁, n₂) is ______.

Options

• 2.00

• 3.00

• 4.00

• 5.00

Answer 1

If two circles `x^2 + y^2 + 2n_1x + 2y + 1/2` = 0 and `x^2 + y^2 + n_2x + n_2y + n_1 = 1/2`, intersect each other orthogonally where n₁, n₂ ∈ I, then number of possible of ordered pairs (n₁, n₂) is 2.00.

Explanation:

2(g₁g₂ + f₁f₂) = c₁ + c₂

⇒ `2(n_1(n_2/2) + (1)(n_2/2))` = n₁

⇒ n₁n₂ + n₂ = n₁

⇒ n₂ = `n_1/((1 + n_1))`

⇒ n₂ = `1 - 1/((1 + n_1))` 

1 + n₁ = 1 or 1 + n₁ = –1

n₁ = 0 or n₁ = –2

⇒ n₂ = 0 or n₂ = 2

The number of ordered pairs (n₁, n₂) is 2 i.e., (0, 0) and (–2, 2)