AITS PT2

x²+y²-4y+3+λx=0
S+λL=0
THE CIRCLE PASSES THROUGH THE POINTS OF INTERSECTION OF S=0
AND L=0
THE FOR BEING OF MINIMUM AREA THE CIRCLE MUST HAVE THE LINE X=0 AS THE DIAMETER
HENCE THE REQUIRED CIRCLE IS X²+Y²-4Y+3=0
RADIUS=√(4-3)=1
RADIUS OF DIRECTOR CIRCLE IS √(1+1)=√2
NO INTERCEPT MADE FROM X AXIS
AND THE LINE \left|2X \right|=1 CUTS THE CIRCLE IN 2 POINTS
HENCE THE CORRECT OPTIONS ARE a) and b)
i know that option d) is also given correct but it is wrong

Thanks Sinchan