Let $$C$$ be the circle with centre $$(0, 0)$$ and radius $$3$$ units. The equation of the locus of the mid points of the chords of the circle $$C$$ that subtend an angle of $$\dfrac{2\pi}{3}$$ at its centre is

Let the given circle be $$C : x^2 + y^2 = 3^2$$ with centre $$O\,(0,0)$$ and radius $$R = 3$$.

Choose any chord $$AB$$ of $$C$$ which subtends a central angle $$\angle AOB = \dfrac{2\pi}{3}$$. Let $$M$$ be the midpoint of this chord. We need the locus of $$M$$.

Step-1 : Relate $$OM$$ to the fixed angle.
The triangle $$\triangle AOB$$ is isosceles with $$OA = OB = R$$ and vertex angle $$\angle AOB = 2\alpha$$, where $$2\alpha = \dfrac{2\pi}{3} \;\Longrightarrow\; \alpha = \dfrac{\pi}{3}.$$ Because $$\triangle AOB$$ is isosceles, the median from $$O$$ to the base $$AB$$ is also the altitude and the angle-bisector. Hence $$OM$$ is perpendicular to $$AB$$ and bisects $$\angle AOB$$, making $$\angle AOM = \alpha = \dfrac{\pi}{3}.$$

Step-2 : Compute the length $$OM$$.
In the right triangle $$\triangle OMB$$, we have
$$OB = R,\quad BM = \dfrac{AB}{2} = R\sin\alpha.$$ Using Pythagoras: $$OM = \sqrt{OB^{2} - BM^{2}} = \sqrt{R^{2} - R^{2}\sin^{2}\alpha} = R\cos\alpha.$$ Substituting $$R = 3$$ and $$\alpha = \dfrac{\pi}{3}$$ gives $$OM = 3\cos\left(\dfrac{\pi}{3}\right) = 3 \times \dfrac{1}{2} = \dfrac{3}{2}.$$

Step-3 : Write the locus.
The midpoint $$M$$ is always at a constant distance $$\dfrac{3}{2}$$ from the origin, so the locus is the circle centred at $$O(0,0)$$ with radius $$\dfrac{3}{2}$$. Its equation is $$x^2 + y^2 = \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}.$

Therefore, the required locus is Option D which is: $$x^2 + y^2 = \dfrac{9}{4}$$.