4Thanks Sinchan
A circle S=0 passes thru the common pts of intersection of family of circles x^2+y^2+λx-4y+3=0
& have min area then
(A) area of S=0 is Î sq units
(B) radius of director circle of S=0 is √2
(C) length of intercept made by S=0 from x-axis is 1 unit
(D) S=0 never cuts l 2x l = 1
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UP 0 DOWN 0 0 2
2 Answers
1x2+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