If the radius of the largest circle with centre ( 2, 0 ) inscribed...

Given equation of ellipse is $\dfrac{{x}^{2}}{36} + \dfrac{{y}^{2}}{9} = 1$

It is of the form $\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1$

With $a = 6, b = 3$.

The centre of the circle is $C\left(2, 0\right)$.

Let $P\left(acos\theta, bsin\theta\right)$ be the common point for ellipse and circle.

$P ≡ \left(6cos\theta, 3sin\theta\right)$

Equation of normal to the ellipse will be $\dfrac{ax}{cos\theta} - \dfrac{by}{sin\theta} = {a}^{2} - {b}^{2}$.

$⇒ 6xsec \theta - 3ycosec \theta = 36 - 9$

But the normal through $P$ passes through the centre of the circle $C\left(2, 0\right)$.

$⇒ 6\left(2\right)sec \theta - 3\left(0\right)cosec \theta = 27$

$⇒ sec \theta = \dfrac{27}{12}$

$⇒ cos\theta = \dfrac{4}{9}$ and $sin\theta = \dfrac{\sqrt{65}}{9}$

$P ≡ \left(6 \times \dfrac{4}{9}, 3 \times \dfrac{\sqrt{65}}{9}\right)$

$P ≡ \left(\dfrac{8}{3}, \dfrac{\sqrt{65}}{3}\right)$

Now $CP = r = \sqrt{{\left(2 - \dfrac{8}{3}\right)}^{2} + {\left(0 - \dfrac{\sqrt{65}}{3}\right)}^{2}}$

$⇒ {r}^{2} = \dfrac{69}{9}$

$⇒ 12{r}^{2} = 12 \times \dfrac{69}{9} = 92$

Therefore, the required answer is $92$