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Just want to note that I've seen a lot of people throwing around the "rule" that if a triangle is inscribed in a circle, then it either a) must be a right triangle or b) have it's hypotenuse pass through the circle's center. Neither of these are true...
The rule is that IF a right triangle is inscribed in a circle, then it's hypotenuse passes through the center of the circle (and is thus a diameter). You can inscribe a virtually infinite number of triangles in a circle which would not be right triangles and would thus not pass through the circle's center.
So to do this problem you must first determine that the triangle formed by the points is indeed a right triangle. Using (Y1-Y2)/(X1-X2) to find the slopes of the three line segments, you will see that segments intersecting at (2,5) have inverse and opposite slopes (3, -1/3) and are thus perpendicular to each other and meet at a right angle. From there it is as simple as applying the pythagorean theorm to find the long side, in this case x^2 = 4^2 + 2^2.
It seems like some people may have gotten "lucky" assuming that the long side was the hypotenuse of a right triangle, and thus a diameter (and the answer) - Just wanted to note that wouldn't always be the case