I believe the answer is (a) but I am not totally sure.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
What you really want to know is, given (1) or (2), does the above equation definitely have a solution at x=0 (the y intercept)
For (1), let x=0. You have
x^2 + y^2 - 2yb + b^2 = 16
y^2 - 2yb = 16 - (a^2 + b^2)
since a^2 + b^2 = 16 from (1), we know that the right hand side of that equation has to be negative. In other words
y^2 - 2yb = some number less than zero
y^2 - 2yb < 0
y^2 < 2yb
either
y < 2b if y is positive
y > 2b if y is negative
There is a definite solution for these two possibilities.
To check for (2), observe the 2 situations
a = b+5 if b is positive
a = 5-b if b is negative
(b+5)^2 + (y-b)^2 = 16
(y-b)^2 = 16 - (b+5)^2
This obviously has a solution when b=-1 because that reduces to
(y+1)^2 = 0, which y =-1
But note that this has no solution when 16-(b+5)^2 < 0 because (y-b)^2 will always be a positive number. So this does NOT always yield the same answer. Thus (2) does not provide enough information.
