Is xy<=1/2?

(1) x^2+y^2=1

The maximum value of xy is when x = y.
x^2+y^2=1
x^2+x^2=1
x = y = 1/√2 and xy = 1/2
The minimum value of xy is when x = -y
x^2+y^2=1
x^2+x^2=1
x = -y = 1/√2 and xy = -1/2
-1/2<xy<1/2. So, statement 1 is sufficient to answer the question(answer = YES).

(2) x^2-y^2=0

(x-y)*(x+y) = 0
x = y or x = -y.
Case 1: If x = y.
Is xy <1/2 ?
Is x^2 <1/2 ?
Let the value of x be 1/2 then x^2 <1/2
Let the value of x be 4 then x^2 >1/2. No definite answer.(You can continue with the case x = -y, but since the first case itself showed that this statement is insufficient to answer the question, I will leave it for you do it.)

Case 2 : x = -y. Please go ahead and solve for this condition.

IMO A