Packed Ellipse

If $\angle AOB = 2\theta$ , then $A,O,B$ can be given coordinates $(-2\sin\theta,0)$ , $(0,-2\cos\theta)$ and $(2\sin\theta,0)$ respectively. Then the ellipse has semi-minor axis $b = 1+2\cos\theta$ , and has equation $\frac{x^2}{a^2} + \frac{y^2}{(1 + 2\cos\theta)^2} \; = \; 1$ The circle $(x - 2\sin\theta)^2 + y^2 \; = \; 1$ must touch the ellipse at exactly two points, so that equation $\begin{aligned} \frac{x^2}{a^2} + \frac{1 - (x - 2\sin\theta)^2}{(1 + 2\cos\theta)^2} & = \; 1 \\ \big[(1 + 2\cos\theta)^2 - a^2\big]\left(x + \frac{2a^2\sin\theta}{(1 + 2\cos\theta)^2-a^2}\right)^2 & = \; \frac{4a^2(1 + \cos\theta)}{(1 + 2\cos\theta)^2 - a^2}\big[(1 + 2\cos\theta)^2 - a^2\cos\theta\big] \end{aligned}$ has exactly one solution in $x$ , which means that $a \; = \; \frac{1 + 2\cos\theta}{\sqrt{\cos\theta}}$ The ellipse and the circle so defined will in fact only meet if $(1+2\sin\theta)\cos\theta < 1$ , or $\theta > 1.21556$ . The diagram, as drawn, is only possible for sufficiently large values of $\theta$ . Simple calculus then shows that the ellipse area $A \; =\; \pi ab \; = \; \pi \frac{(1 + 2\cos\theta)^2}{\sqrt{\cos\theta}}$ is minimized when $\cos\theta = \tfrac16$ , and hence when $\tan\theta = \sqrt{35}$ . Thus the desired answer is $\boxed{2\tan^{-1}\sqrt{35}}$ .

Let the circles be unit circles, let the ellipse be centered at the origin $C$ , and let $\angle AOB = 2\theta$ . Also, let $D$ be one of the points of tangency between the ellipse and circle $B$ , and $E$ be the point of tangency between the ellipse and circle $O$ , as shown below.

Then $\angle BOC = \theta$ , and $BC = 2 \sin \theta$ and $CO = 2 \cos \theta$ . This means the minor axis of the ellipse $b = CE = 2 \cos \theta + 1$ and $B$ has coordinates $(2 \sin \theta, 0)$ .

Let $D$ have coordinates $(p, q)$ and let $a$ be the major axis of the ellipse. Since $D$ is on the ellipse, $\frac{p^2}{a^2} + \frac{q^2}{b^2} = 1$ . Since $D$ is on circle $B$ , $(p - 2 \sin \theta)^2 + q^2 = 1$ . Since the slope of $BD$ must be the same as the slope of the normal of the ellipse at $(p, q)$ , $\frac{q}{p - 2 \sin \theta} = \frac{a^2q}{b^2p}$ . Solving these $3$ equations, along with the identity $\sin^2 \theta + \cos^2 \theta = 1$ , we find that $a$ reduces to $a = b\sqrt{\frac{2}{b - 1}}$ .

The area of an ellipse is $A = \pi ab$ , so in this ellipse, $A = \pi b^2 \sqrt{\frac{2}{b - 1}}$ . Using a derivative we find that this area is minimized when $b = \frac{4}{3}$ . Since $b = 2 \cos \theta + 1$ , $\cos \theta = \frac{1}{6}$ , which makes $\tan \theta = \sqrt{35}$ , and $\angle AOB = \boxed{2 \tan^{-1} \sqrt{35}}$ .

This is simpler way of doing this problem using the options given.

Consider the half triangle of the bigger triangle AOB and take the corresponding sides of the triangle as $x$ and $y$

Then by Pythagoras Theorem,

$x^{2}+y^{2} = 4r^{2}$ where $r$ is the radius of the individual circles.

Now, we know that, $\tan( AOB/2) = \dfrac {x}{y}$

So the following conditions should satisfy-

$1]x^{2}+y^{2}$ should be divisible by $4$ and

$2]$ The divisible number $\dfrac {x^2+y^2}{4}$ should be a perfect square number. Check all options and only the last one satisfies.

$ANSWER :\boxed{2\tan^{-1}\sqrt{35}}$

$Note:$ This method works only if you consider $r$ to be an $Integer$

Answer: Using Pictorial Reference, James Buddenhagen, June 2004, and plotting with WolframAlpha the following is obtained as the answer:

Ellipse with

semi-major axis=8/√6,

semi-minor axis=4/3,

(1) Center Unit Circle = ( 0, -1/3),

(2) Right Unit Circle = ((1/3)√35,0),

(3) Left Unit Circle. = (-(1/3)√35,0)