Let C be the largest circle centred at (2, 0) and inscribed in the ellipse $\frac{{\mathrm{x}}^2}{36}+\frac{{\mathrm{y}}^2}{16}=1$. If (1, $\mathrm{\alpha }$) lies on C, then 10$\mathrm{\alpha }$²is equal to ________________

Let circle is $(\mathrm{x}-2{)}^2+{\mathrm{y}}^2={\mathrm{r}}^2,$

$\frac{{\mathrm{x}}^2}{36}+\frac{{\mathrm{y}}^2}{16}=1\Rightarrow \frac{{\mathrm{x}}^2}{36}+\frac{{\mathrm{r}}^2-(\mathrm{x}-2{)}^2}{16}=1$
$\Rightarrow 16{\mathrm{x}}^2+36\left({\mathrm{r}}^2-(\mathrm{x}-2{)}^2\right)=\left(36\right)\left(16\right)$
Using $∆$ = 0 we get ${\mathrm{r}}^2=\frac{320}{25},\text{ circle is }(\mathrm{x}-2{)}^2+{\mathrm{y}}^2=\frac{320}{25}$
(1, $\mathrm{\alpha }$) lies on this circle $\Rightarrow {\mathrm{\alpha }}^2=\frac{59}{5}\therefore \boxed{10{\mathrm{\alpha }}^2=118}$