Let the orthocentre and centroid of a triangle be A(-3, 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:

We have the orthocentre $$H\equiv A(-3,\,5)$$ and the centroid $$G\equiv B(3,\,3)$$ of an unknown triangle. Let the circumcentre be $$O\equiv C(x,\,y).$$ In every triangle, the three special points $$H,\;G,\;O$$ lie on a single straight line called the Euler line. A well-known property of the Euler line is the ratio

$$HG:GO = 2:1.$$

This means that the centroid $$G$$ divides the segment joining the orthocentre $$H$$ and the circumcentre $$O$$ internally in the ratio $$2:1,$$ the longer part being adjacent to $$H$$.

If a point $$P(x_P,\,y_P)$$ divides the segment joining $$A(x_1,\,y_1)$$ and $$B(x_2,\,y_2)$$ internally in the ratio $$m:n$$ (with $$AP:PB = m:n$$), then the section formula gives

$$P\;=\;\left(\dfrac{n x_1 + m x_2}{m+n},\; \dfrac{n y_1 + m y_2}{m+n}\right).$$

Here $$A\equiv H(-3,\,5),\; B\equiv O(x,\,y),\; P\equiv G(3,\,3)$$ and the ratio is $$m:n = 2:1.$$ Therefore

$$G\;=\;\left(\dfrac{1\cdot(-3) + 2\cdot x}{1+2},\; \dfrac{1\cdot 5 + 2\cdot y}{1+2}\right).$$

But the coordinates of $$G$$ are already known to be $$(3,\,3).$$ Equating the corresponding components, we obtain two linear equations:

$$\dfrac{-3 + 2x}{3} = 3 \quad\text{and}\quad \dfrac{5 + 2y}{3} = 3.$$

Clearing the denominators by multiplying each equation by $$3$$ gives

$$-3 + 2x = 9 \quad\text{and}\quad 5 + 2y = 9.$$

Now, solving for $$x$$ and $$y$$ one by one:

$$2x = 9 + 3 = 12 \;\;\Longrightarrow\;\; x = \dfrac{12}{2} = 6,$$

$$2y = 9 - 5 = 4 \;\;\Longrightarrow\;\; y = \dfrac{4}{2} = 2.$$

So the circumcentre is located at

$$O\equiv C(6,\,2).$$

We must now find the length of segment $$AC$$ because the required circle has $$AC$$ as its diameter. Using the distance formula between points $$A(x_1,\,y_1)$$ and $$C(x_2,\,y_2):$$

$$AC \;=\; \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$$

Substituting $$A(-3,\,5)$$ and $$C(6,\,2)$$ gives

$$AC \;=\; \sqrt{\bigl(6 - (-3)\bigr)^2 + \bigl(2 - 5\bigr)^2}$$ $$\;\;=\; \sqrt{9^2 + (-3)^2}$$ $$\;\;=\; \sqrt{81 + 9}$$ $$\;\;=\; \sqrt{90}$$ $$\;\;=\; 3\sqrt{10}.$$

Because $$AC$$ is the diameter of the required circle, its radius $$r$$ is exactly half of $$AC$$:

$$r \;=\; \dfrac{AC}{2} \;=\; \dfrac{3\sqrt{10}}{2}.$$

To match the options, notice that

$$3\sqrt{\dfrac{5}{2}} \;=\; 3\,\dfrac{\sqrt{5}}{\sqrt{2}} \;=\; 3\,\dfrac{\sqrt{10}}{2} \;=\; \dfrac{3\sqrt{10}}{2}.$$

Thus the numerical value we obtained coincides with Option D.

Hence, the correct answer is Option 4.