The centroid, orthocentre, and circumcentre of any triangle always lie on a straight line known as the Euler line, and the centroid divides the distance from the orthocentre to the circumcentre in the ratio 2:1. Let the orthocentre be A(-3, 5) and the centroid be B(3, 3). Using the section formula, the coordinates of the circumcentre C(x, y) can be found since B divides AC in the ratio 2:1. This gives the equations 3 = (2x - 3)/3 and 3 = (2y + 5)/3. Solving the first equation yields 9 = 2x - 3, so 2x = 12 and x = 6. Solving the second equation yields 9 = 2y + 5, so 2y = 4 and y = 2. The circumcentre C is at (6, 2). The question asks for the radius of the circle having the line segment AC as its diameter. We calculate the distance between A(-3, 5) and C(6, 2), which is sqrt((6 - (-3))^2 + (2 - 5)^2) = sqrt(9^2 + (-3)^2) = sqrt(81 + 9) = sqrt(90). The radius is half of this diameter, so we divide by 2 to get sqrt(90)/2. Simplifying the fraction under the square root gives sqrt(90/4) = sqrt(45/2) = 3*sqrt(5/2). The result is 3*sqrt(5/2).