If the tangent at a point $(a \cos \theta , b \sin \theta )$ on the ellipse $x^2/a^2+y^2/b^2=1$ meets the auxiliary circle in two points, the chord joining them subtends a right angle at the centre  then the eccentricity of the ellipse is given by

Equation of the tangent $(a\cos \theta ,b\sin \theta )$ to the ellipse $x^2/a^2+y^2/b^2=1$is

$\Rightarrow \frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$         (I)

The joint equation of the lines joining the points of intersection of (i) and the auxillary circle $x^2+y^2=a^2$ to the origin, which is the centre of the circle, is

$x^2+y^2=a^2{\left[\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta \right]}^2$ 

Since these lines are at right angles Co-efficient of ${\mathrm{x}}^2+$ Co-efficient of $y^2=0$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1-a^2\left(\frac{{\cos }^2\theta }{a^2}\right)+1-a^2\left(\frac{{\sin }^2\theta }{b^2}\right)=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\sin }^2\theta \left(1-\frac{a^2}{b^2}\right)+1=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\sin }^2\theta \left(b^2-a^2\right)+b^2=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\sin }^2\theta \left[a^2\left(1-e^2\right)-a^2\right]+a^2\left(1-e^2\right)=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(1+{\sin }^2\theta \right)a^2{e}^2=a^2\Rightarrow e={\left(1+{\sin }^2\theta \right)}^{-1/2}\end{array}$