Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
$$
x^{2}=a^{2}-\frac{a^{2} y^{2}}{b^{2}} \ldots . \cdots(i)
$$
The equation of circle will be $x^{2}+(y-b)^{2}=1$....... (ii)
Using (i) in (ii), we get
$$
\begin{array}{l}
a^{2}-\frac{a^{2} y^{2}}{b^{2}}+(y-b)^{2}=1 \\
\left(1-\frac{a^{2}}{b^{2}}\right) y^{2}+2 b y+a^{2}+b^{2}-r^{2}=0
\end{array}
$$
As you can see from the figure the ellipse touches the circle for same value of $y$ so the roots of the eqaution are equal $\therefore \mathrm{D}=0$ $4 \mathrm{~b}^{2}-4\left(1-\frac{\mathrm{a}^{2}}{\mathrm{~b}^{2}}\right)\left(\mathrm{a}^{2}+\mathrm{b}^{2}-\mathrm{r}^{2}\right)=0$ $\mathrm{r}^{2}=\frac{\mathrm{a}^{4}}{\mathrm{a}^{2}-\mathrm{b}^{2}}$ $\Rightarrow \mathrm{b}=\mathrm{a} \sqrt{1-\frac{\mathrm{a}^{2}}{\mathrm{r}^{2}}}$ Area of ellipse $=\pi \mathrm{ab}$ $\mathrm{A}=\pi \times \mathrm{a} \times \mathrm{a} \sqrt{1-\frac{\mathrm{a}^{2}}{\mathrm{r}^{2}}}$
As area of ellipse is maximun
$$
\begin{array}{l}
\therefore \frac{\mathrm{d} \mathrm{A}}{\mathrm{d} \mathrm{a}}=0 \\
\Rightarrow \mathrm{a}^{2}=\frac{2 \mathrm{r}^{2}}{3} \\
\mathrm{a}=\sqrt{\frac{2}{3}} \mathrm{r} \\
\therefore \mathrm{b}=\mathrm{a} \sqrt{1-\frac{2}{3}}=\frac{\mathrm{a}}{\sqrt{3}} \\
\mathrm{e}^{2}=1-\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}} \\
\Rightarrow \mathrm{e}=\sqrt{\frac{2}{3}}
\end{array}
$$