There is an alternate solution to it that does not need any calculation.
Note that the diameter is the biggest chord in the circle and as the perpendicular distance of chord increases from center, the chord becomes shorter. Therefore, chord XY is bigger as it has lesser perpendicular distance as compared to chord PQ.
Hence answer is B.

Note that both triangles OPQ and OXY are both isosceles triangles (OP=OQ=OX=OY=radius of the circle) => the height drawn from O to either bases cut the chord PQ and XY in equal halt => by Pythagorean theorem: radius^2 = (5.9)^2+(PQ/2)^2 = (5.8)^2+(XY/2)^2. Since (5.9)^2>(5.8)^2 => (PQ/2)^2<(XY/2)^2 => PQ<XY (PQ and XY are positive of course) => B is answer.

You could go on full length to prove this but there's just one thing to remember here: The closer to the center, the longer the chord (here we talk about the distance between the center and the point on the chord that's closest to the center)

When it's the furthest from the center it's 0 and when it's closest the length becomes the diameter.

XY is closer to the center so XY is longer.
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Nearer chord to diameter is larger.
Answer B

how do we infer that the perpendicular from center of circle to PQ or XY cuts it into half?

There's a nice/important triangle property that says: The altitude of an isosceles (or an equilateral) triangle bisects the base


(screenshot from my course)

Since OX and OY are both radii of the same circle, we can conclude that OX = OY.
So, following the above property, we know that the altitude cuts XY in half (i.e., bisects XY)

Cheers,
Brent
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The simplest way of doing this is noticing that the line from the center O is perpendicular to XY and OY. Thus, 5.9 and 5.8 are measure the distance of the (center of the) chord from the center of the circle. Now the farther the chord is from the center, the smaller will be its length. Therefore, XY will be greater than PQ.

(center of the chord) = because the altitude of an isosceles triangle bisects the base. OX = OY = OP = OQ = radius.
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