Let $\mathrm{C}:{\mathrm{x}}^2+{\mathrm{y}}^2=4$ and ${\mathrm{C}}^{\text{'}}:{\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{\lambda x}+9=0$ be two circles. If the set of all values of $\lambda$ so that the circles C and ${\mathrm{C}}^{\text{'}}$ intersect at two distinct points, is $\mathrm{R}-[\mathrm{a},\mathrm{b}],$ then the point $(8\mathrm{a}+12,16\mathrm{b}-20)$ lies on the curve :

${\mathrm{x}}^2+{\mathrm{y}}^2=4$

$\mathrm{C}(0,0) {\mathrm{r}}_1=2$

${\mathrm{C}}^{\text{'}}(2\mathrm{\lambda },0) {\mathrm{r}}_2=\sqrt{4{\mathrm{\lambda }}^2-9}$

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

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

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

$\text{True }\mathrm{\lambda }\in \text{ R ... (1) }$

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

$\text{True }\mathrm{\lambda }\in \text{ R.... (1) }$

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

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

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

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

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

from (1) and (2) $\mathrm{\lambda }\in$

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

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

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