Let complex numbers  $Z_1$ and  $\frac{1}{\stackrel{-}{Z_1}}$ lie on $|Z-Z_0|=4R$ and $|Z-Z_0|=16R$  respectively. If $Z_0$  satisfies  $2{\left|Z_0\right|}^2=R^2+2$ and ${\left|Z_1\right|}^2=\frac{p}{q}$ where  $p,q,\in N$  and  $H.C.F$ of p and q is 1, then choose the correct option(s).

$$\begin{gathered} (Z_1-Z_0)({\overline{Z}}_1+{\overline{Z}}_0)=16R^2 \\ \Rightarrow Z_1{\overline{Z}}_1-Z_0{\overline{Z}}_1-{\overline{Z}}_0{Z}_1+{\left|Z_0\right|}^2=16R^2 \mathrm{....}\left(1\right) \\ (\frac{1}{Z_1}-Z_0)(\frac{1}{Z_1}-{\overline{Z}}_0)=256R^2 \\ \frac{1}{Z_1{\overline{Z}}_1}-\frac{Z_0}{{\overline{Z}}_1}-\frac{{\overline{Z}}_0}{Z_1}+{\left|Z_0\right|}^2=256R^2 \\ Put {\left|Z_1\right|}^2=k \\ 1-Z_0{\overline{Z}}_1-{\overline{Z}}_0{Z}_1+K{\left|Z_0\right|}^2=256R^{2}K \\ \left(1\right)-\left(2\right) \\ \Rightarrow (K-1)+(1-K){\left|Z_0\right|}^2=16R^2(1-16K) \\ \Rightarrow (1-K)+({\left|Z_0\right|}^2-1)=16R^2(1-16K) \\ \Rightarrow (1-K)+(1-16K) \Rightarrow K=\frac{31}{511} \end{gathered}$$