Pls explain the question LET C BE A CIRCLE ITH CENTRE 0,0 AND RADIUS = 3 UNIT THEN THE EQUATION OF THE LOCUS OF THE MID POINTS OF THE CHORD THAT SUBTEND AN ANGLE = 2PI/3?

Let chord AB subtends an angle $\frac{2\mathrm{\pi }}{3}$ at the centre.
Let OP be the perpendicular bisector of the chord AB as shown below:


$$\begin{gathered} OA=OB=r \\ \angle AOB=\frac{2\mathrm{\pi }}{3} \\ Since, OP is the perpendicular bisector so \\ \angle AOP=\angle POB=\theta \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{3} \end{gathered}$$

The equation of the circle is $x^2+y^2=9$. So, the equation of the chord with mid-point at $P\left(h, k\right)$ will be

$$\begin{gathered} hx+ky-9=h^2+k^2-9 \\ \Rightarrow hx+ky=h^2+k^2\dots \dots \dots \left(i\right) \end{gathered}$$

I $$\begin{gathered} In △OPB \\ \cos \left(\theta \right)=\frac{OP}{OB} where OP=\sqrt{h^2+k^2} and OB =r \left(radius\right) \\ \Rightarrow r\cos \left(\theta \right)=\sqrt{h^2+k^2} \\ \Rightarrow 3\times \cos \left(\frac{\mathrm{\pi }}{3}\right)=\sqrt{h^2+k^2} \\ \Rightarrow 3\times \frac{1}{2}=\sqrt{h^2+k^2} \\ \Rightarrow h^2+k^2=\frac{9}{4}\dots \dots \dots \left(ii\right) \end{gathered}$$

From (i) and (ii) the locus of the mid-point (h, k) will be

$$\begin{gathered} hx+ky=\frac{9}{4} \\ Replace h\to x and k\to y we get \\ {x}^2+y^2=\frac{9}{4} \end{gathered}$$