If a and c are positive real numbers and the ellipse $\frac{{\mathrm{x}}^2}{4{\mathrm{c}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{c}}^2}=1$has four distinct points in the common with circle ${\mathrm{x}}^2+{\mathrm{y}}^2=9{\mathrm{a}}^2,$ then

Given circle  ${\mathrm{x}}^2+{\mathrm{y}}^2=9{\mathrm{a}}^2$

$$\begin{gathered} Given ellipse \frac{{\mathrm{x}}^2}{4{\mathrm{c}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{c}}^2}=1 \\ Given circle x^2+y^2=9a^2 \\ centre=(0,0),r=3a \\ Since circle and ellipse intersects at four distinct points then \\ r>2c and r<c \\ \Rightarrow 3a-2c>0 and 3a-c<0 \\ \Rightarrow (3a-2c)(3a-c)<0 \\ \Rightarrow 9a^2-9ac+2c^2<0 \\ \Rightarrow 9ac-9a^2-2c^2>0 \end{gathered}$$