Question

Q. 87. If two circles \( x^{2}+y^{2}+2 n_{1} x+2 y+\frac{1}{2}=0 \) and \( x^{2}+y^{2}+n_{2} x+n_{2} y+n_{1}=\frac{1}{2}, \quad \) intersect each other orthogonally where \( n_{1}, n_{2} \in I \), then number of possible of ordered pairs \( \left(n_{1}, n_{2}\right) \) is

Answer 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)