The correct options are
A (1,2)
D (−1,−2)
As the line ax+by=0 touches the circle x2+y2+2x+4y=0, so the distance of the centre (−1,−2) from the line is equal to radius
∣∣∣−a−2b√a2+b2∣∣∣=√(−1)2+(−2)2−0
Squaring both the sides, we get
⇒(a+2b)2=5(a2+b2)
⇒4a2−4ab+b2=0
⇒(2a−b)2=0
⇒b=2a
Now, ax+by=0 is a normal to the circle x2+y2−4x+2y−3=0, so it should pass through the centre (2,−1)
⇒2a−b=0
⇒b=2a
Only, (a,b)=(1,2) and (−1,−2) are satisfying the condition.