I received a PM asking me to comment.abhi332 wrote:Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ?
(1) a^2+b^2>16
(2) a=|b|+5
(x-h)² + (y-k)² = r² is the equation of a CIRCLE with a center at (h,k) and a radius of length r.
Thus, in a circle with equation (x-a)² + (y-b)²=16:
Center = (a,b).
r=4.
For the circle to intersect the y-axis, the x value of the center (a) cannot be more than the length of the radius (4).
Question rephrased: Is a≤4?
Statement 1: a²+b²>16
It's possible that a=3 and b=3, yielding the following circle:
It's possible that a=5 and b=5, yielding the following circle:
Since in the first case the circle intersects the y-axis, and in the second case the circle does not intersect the y-axis, INSUFFICIENT.
Statement 2: a=|b|+5
The smallest possible value of |b| is 0.
Thus, the smallest possible value of a is 5, yielding the following circle:
If b is ANY OTHER VALUE, the value of a will INCREASE, moving the circle FARTHER from the y=axis.
To illustrate, if b=-1 so that a=|-1|+5=6, the circle will have a center at (6,-1):
Thus, it is not possible for the circle to intersect the y-axis.
SUFFICIENT.
The correct answer is B.
