$ \begin{aligned} & \text { Let } H: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\left(b^2=a^2\left(e^2-1\right)\right) \\ & \therefore e q^n \text { of } C_1=x^2+y^2=a^2 \\ & \text { Ar. }=36 \pi \\ & \pi a^2=36 \pi \\ & a=6 \\ & \text { Now radius of } C_2 \text { can be } a(e-1) \text { or } a(e+1) \\ & \text { for } r=a(e-1) \\ & \text { for } r=a(e+1) \\ & \text { Ar. }=4 \pi \\ & \pi r^2=4 \pi \\ & \pi a^2(e-1)^2=4 \pi \\ & a^2(e+1)^2=4 \\ & 36 \pi(e-1)^2=4 \pi \\ & 36(e+1)^2=4 \\ & e-1=\frac{1}{3} \\ & e+1=\frac{1}{3} \\ & e=\frac{4}{3} \\ & -\frac{2}{3} \end{aligned} $
Not possible
$\begin{aligned} & \therefore \mathrm{b}^2=36\left(\frac{16}{9}-1\right)=28 \\ & \therefore \mathrm{LR}=\frac{2 \mathrm{~b}^2}{\mathrm{a}}=\frac{2 \times 28}{6}=\frac{28}{3}\end{aligned}$