If two circles touching externally at origin also touch the ellipse $\frac{{\mathrm{x}}^2}{10}+\frac{{\mathrm{y}}^2}{8}=1$ as shown and if the square of the length of their direct common tangent is $\frac{\mathrm{a}}{\mathrm{b}}$ where $\text{ G.C.D }(\mathrm{a},\mathrm{b})=1,\mathrm{a},\mathrm{b}\in \mathrm{N}$ then the value of  a – b² equals

 

$$\begin{gathered} Given ellipse \frac{{\mathrm{x}}^2}{10}+\frac{{\mathrm{y}}^2}{8}=1 \\ and let circles be (\mathrm{x}-\mathrm{r}{)}^2+{\mathrm{y}}^2={\mathrm{r}}^2 \end{gathered}$$  

$\Rightarrow {\mathrm{x}}^2-2\mathrm{rx}+{\mathrm{r}}^2+{\mathrm{y}}^2={\mathrm{r}}^2$
$\begin{array}{l}\Rightarrow {\mathrm{x}}^2-2\mathrm{rx}+8\left(1-\frac{{\mathrm{x}}^2}{10}\right)=0\\ \Rightarrow \frac{1}{5}{\mathrm{x}}^2-2\mathrm{rx}+8=0\text{ for which }\mathrm{\Delta }=0\\ \Rightarrow 4{\mathrm{r}}^2=4\left(\frac{1}{5}\right)\left(8\right)\Rightarrow 4{\mathrm{r}}^2=\frac{32}{5}\end{array}$
$\therefore$Square of length of direct common tangent $=(2\mathrm{r}{)}^2=\frac{32}{5}$

$$\begin{gathered} Given that \frac{a}{b}=\frac{32}{5} and G.C.D (a,b)=1 \\ So,a=32,b=5 \\ Now a-b^2=32-25=7 \end{gathered}$$