ICOT: This riddle became popular in the early 1930s, and if it were before 1952, you would not only have the correct answer, but you also noted within your answer the correct strategy X would use to catch Y.
Namely, X could be guaranteed to catch Y by staying on the same radius as Y. In other words, X moves at top speed in such a way that X always lies on the radius vector from the center to Y. If we assume without loss of generality that X stays on the boundary of the circle, then it is easy to check that X does now catch Y in
finite time. Indeed, in the time that it takes Y to run a quarter-circle at full speed, X (say starting at the center) performs a semicircle of half the radius of the disc, thus catching Y.
However, in 1952, it was proven that Y need not stay on the boundary and in fact can survive forever in a closed circle. The clever proof is as follows:
First, I establish a winning strategy for Y, and then I will establish that it is feasible, i.e. Y does not leave the closed circle.
Let T = {t_1, t_2, t_3, …} be a sequence of time periods of equal length. At the ith time period, Y runs in a straight line that is perpendicular to Y’s radius vector at the start of the period. Y runs into the half-plane that does not contain X (if X is on the radius then Y may choose either half-plane). Certainly, X does not catch Y in this time period.
Let Y repeat this procedure for all time periods. As long as T is an infinite sequence, Y is never caught.
Now, I must show that this strategy is feasible. Note that if R_i is the distance of Y from the center at the start of the ith time period, then (R_i+1)^2 = (R_i)^2 + (t_i)^2. (This is just a simple application of the famous Pythagorean Theorem.) If the sum of (t_i)^2 is finite, then the R_i’s are bounded so that (multiplying by a constant if necessary) Y does not leave the closed circle. Let t_i = 1/i.
Then, Y is never caught and never leaves the closed circle.