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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