Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ?
(1) a^2+b^2>16
(2) a=|b|+5
Answer: B
The given curve will intersect the y-axis when x=0
Thus we get a^2 + (y-b)^2 = 16
<=> a^2 + y^2 + b^2 - 2yb = 16
<=> y^2 - 2yb + a^2 + b^2 -16 = 0
In order to have real roots b^2 - 4ac >= 0
=> 4b^2 - 4(1)(a^2 + b^2 -16) >=0
=> a^2 <=16
From statement (1): Given that a^2 + b^2 > 16
No information about a^2 <=16 --- hence insufficient
From statement (2): Given a = |b|+5
=> |b| is positive or zero.
=> a is atleast equal to 5 and value of a^2 is atleast 25.
But we know that a^2 <=16 ====> does not intersect the y axis ---- hence sufficient
Hence the answer B
Wednesday, June 04, 2008
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