The centre of the circle having end points of its one diameter as (-4, 2) and (4, -3) is:

1. (0, -1)

2. (2, -1)

3. $\Big(0, -\dfrac{1}{2}\Big)$

4. $\Big(4, -\dfrac{5}{2}\Big)$

Let the end points of the diameter be A(-4, 2) and B(4, -3). The centre of the circle is the mid-point of the diameter AB.

By mid-point formula,

(x, y) = $\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

Substituting values we get :

$\Rightarrow (x, y) = \Big(\dfrac{-4 + 4}{2}, \dfrac{2 + (-3)}{2}\Big) \\[1em] = \Big(\dfrac{0}{2}, \dfrac{-1}{2}\Big) \\[1em] = \Big(0, -\dfrac{1}{2}\Big).$

Hence, Option 3 is the correct option.