If the radius of the largest circle with centre $(2,0)$ inscribed in the ellipse ${\mathrm{x}}^2+4{\mathrm{y}}^2=36$ is r, then $12{\mathrm{r}}^2$ is equal to

Circle ${(\mathrm{x}-2)}^2+{\mathrm{y}}^2={\mathrm{r}}^2$

${(\mathrm{x}-2)}^2+9(1-\frac{{\mathrm{x}}^2}{36})={\mathrm{r}}^2$
$4{(\mathrm{x}-2)}^2+36-{\mathrm{x}}^2=4$

$r^{2}3{\mathrm{x}}^2-16\mathrm{x}+52-4{\mathrm{r}}^2=0$

$\mathrm{\Delta }=0\Rightarrow 256-12(52-4{\mathrm{r}}^2)=0$
$16-3(13-{\mathrm{r}}^2)=0$

$3{\mathrm{r}}^2=23\Rightarrow 12{\mathrm{r}}^2=92$