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

Given equation of ellipse is $\frac{{x}^{2}}{36}+\frac{{y}^{2}}{9}=1$36x2​+9y2​=1

It is of the form $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$a2x2​+b2y2​=1

With $a=6,b=3$a=6,b=3.

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

Let $P(a\mathrm{cos}\theta ,b\mathrm{sin}\theta )$P(acosθ,bsinθ) be the common point for ellipse and circle.

$$P\equiv (6\mathrm{cos}\theta ,3\mathrm{sin}\theta )$$P≡(6cosθ,3sinθ)

Equation of normal to the ellipse will be $\frac{ax}{\mathrm{cos}\theta }-\frac{by}{\mathrm{sin}\theta }={a}^{2}-{b}^{2}$cosθax​−sinθby​=a2−b2.

$$\Rightarrow 6xsec\theta -3ycosec\theta =36-9$$⇒6xsecθ−3ycosecθ=36−9

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

$$\Rightarrow 6(2)sec\theta -3(0)cosec\theta =27$$⇒6(2)secθ−3(0)cosecθ=27

$$\Rightarrow sec\theta =\frac{27}{12}$$⇒secθ=1227​

$\Rightarrow \mathrm{cos}\theta =\frac{4}{9}$⇒cosθ=94​ and $\mathrm{sin}\theta =\frac{\sqrt{65}}{9}$sinθ=965​​

$$P\equiv (6\times \frac{4}{9},3\times \frac{\sqrt{65}}{9})$$P≡(6×94​,3×965​​)

$$P\equiv (\frac{8}{3},\frac{\sqrt{65}}{3})$$P≡(38​,365​​)

Now $CP=r=\sqrt{{(2-\frac{8}{3})}^{2}+{(0-\frac{\sqrt{65}}{3})}^{2}}$CP=r=(2−38​)2+(0−365​​)2​

$$\Rightarrow {r}^{2}=\frac{69}{9}$$⇒r2=969​

$$\Rightarrow 12{r}^{2}=12\times \frac{69}{9}=92$$⇒12r2=12×969​=92

Therefore, the required answer is $92$92