We have OP, hypotenuse of triangle PZO and radius of Circle, equal to 2. Since POQ is 90` and O is the origin of Circle and PO=OQ, angle OPZ(coordinates of Z [x=0;y=1])=OQZ=45`. These two angles are the angles of the base of isosceles triangle OPQ (PO=OQ).
So sorry ---> Q coordinate can't be always symmetric to P coordinate [Sqrt(3);1]
triangle POQ is an equilateral right triangle, where PQ=2*Sqrt(2); If the Coordinates P and Q are symmetric then 2*Sqrt(2)/2 and s=Sqrt(2) otherwise following is effected...
... It can be the case, when the triangle is shifted left or right (2 pic below/attached) To learn when our points in P, Q coordinates are symmetric, skewed to the left or right we need to estimate the angles PO-abscess x and QO-abscess x; if these two angles are 45`, then we have symmetry - otherwise we have skewness

And we have angle relationship (30-60-90 for Sqrt(3)-1-2 sides) with 30` POX1 (at the left side), so our triangle POQ is skewed to the left. But we know that the distance between P and Q is always 2Sqrt(2) because of right anlge POQ. We need only to find the projection of line PQ

angle QOX=60` (180-90-30) and we have the right triangle relationship (30-60-90) where hypotenuse is 2 and s-coordinate for Q is opposite to 30` and must equal to 1, whereas t-coordinate of Q is Sqrt(3)

So the answer s=1 and I am as sure as I have exam on the 17th of February that the correct answer must be B

RACHVIK wrote:Pls find attached question from GMAT Prep. OA after some comments later in the day as I feel the answer is incorrect in software.

final attachment

Thanks Night Reader. The OA is B.