The correct option is D (−√2,−1)∪(1,√2)
Given equation of circle is x2+y2=1
C≡(0,0) and r=1
Angle between pair of tangents is given by
θ=2tan−1(r√S1)
√S1=√a2+02−1
⇒√S1=√a2−1 ⋯(1)
For (1) to be defined,
a2−1≥0
⇒(a+1)(a−1)≥0
⇒a∈(−∞,−1]∪[1,∞) ⋯(2)
Now, θ=2tan−11√a2−1
Since π2<θ<π,
⇒π2<2tan−11√a2−1<π1
⇒π4<tan−11√a2−1<π2
⇒1<1√a2−1<∞
⇒0<√a2−1<1
⇒0<a2−1<1
⇒a2−2<0 and a2−1>0
⇒(a+√2)(a−√2)<0 and (a+1)(a−1)>0
⇒a∈(−√2,√2) and a∈(−∞,−1)∪(1,∞)
⇒a∈(−√2,−1)∪(1,√2) ⋯(3)
On taking intersection of (2) and (3), we get
a∈(−√2,−1)∪(1,√2)