Let the orthocentre and centroid of a triangle be A(-7,5) and B(3,3) respectively. If C is the circumcentre of this triangle, then the radius of the circle having line segment AC as diameter, is

B (centroid ) divides orthocentre and circumcentre in ratio 2:1
Let's circumcentre be (c) = (x,y)
(3,3); B = 2(x) + 1(-7)/3, 2(y) + 1(5)/3
2x - 7 = 9 => x = 6
2y + 5 = 9 => y = 2
Circumcentre will be (6,2)
AC is a diameter .
A(-7,5) C(6,2)
AC = √9² + 3² = √90
Radius = AC/2 = √90/2 = √90/√4 = √(90/4) = √(45/2) = 3√5/√2 = 3√5√2/2 = 3√10/2
The question in the input had options as 3√52,√10,2√10. None of which matches with 3√10/2. However, if we assume that the centroid divides the orthocenter and circumcenter in the ratio 1:2, we proceed as follows:
Let the orthocenter be H(-7,5) and the centroid be G(3,3). Let the circumcenter be C(x,y).
Then, using the section formula, we have:
3 = (1x + 2(-7))/(1+2) = (x-14)/3
=> x = 23
3 = (1y + 25)/(1+2) = (y+10)/3
=> y = -1
Therefore, the circumcenter is C(13,-1).
Then AC = √((13+7)² + (-1-5)²) = √(400 + 36) = √436
Radius = AC/2 = √436/2 = √109
This is not amongst options either. It appears there's some inconsistency or error either in the question or the solution.