$x^2+y^2=4$

$C(0,0) \quad \quad r_1=2$

$C^{\prime}(2 \lambda, 0) \quad r_2=\sqrt{4 \lambda^2-9}$

$\left|\mathrm{r}_1-\mathrm{r}_2\right|<\mathrm{CC} \mathrm{C}^{\prime}<\left|\mathrm{r}_1+\mathrm{r}_2\right|$

$\left|2-\sqrt{4 \lambda^2-9}\right|<|2 \lambda|<2+\sqrt{4 \lambda^2-9}$

$4+4 \lambda^2-9-4 \sqrt{4 \lambda^2-9}<4 \lambda^2$

True$\lambda \in$ R…. $(1)$

$4 \lambda^2<4+4 \lambda^2-9+4 \sqrt{4 \lambda^2-9}$

$5<4 \sqrt{4 \lambda^2-9} \text { and } \quad \lambda^2 \geq \frac{9}{4}$

$\frac{25}{16}<4 \lambda^2-9 \quad \lambda \in\left(-\infty,-\frac{3}{2}\right] \cup\left[\frac{3}{2}, \infty\right)$

$\frac{169}{64}<\lambda^2$

$\lambda \in\left(-\infty,-\frac{13}{8}\right) \cup\left(\frac{13}{8}, \infty\right)$       $…(2)$

from $(1) $and $(2)$$\lambda \in$

$\lambda \in\left(-\infty,-\frac{13}{8}\right) \cup\left(\frac{13}{8}, \infty\right) \Rightarrow R-\left[-\frac{13}{8}, \frac{13}{8}\right]$

as per question $a=-\frac{13}{8}$ and $b=\frac{13}{8}$

$\therefore \quad$ required point is $(-1,6)$ with satisfies option $(4)$