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2 circles ($r_1 \neq r_2$) and one ellipse touch each other as shown in Figure-1. What is the minimum area (A) among them ? Please consider $a,b,r_1,r_2$ given values(constants). Let's imagine we change positions of circles and the ellipse to get minimum area (A) among them without changing $a,b,r_1,r_2$ values.

I believe I found the strategy how can be solved for ($r_1 = r_2$) condition as shown in Figure-2. But I haven't got a way for ($r_1 \neq r_2$) condition.

Thanks for help and answers.! enter image description here

EDIT:

I have drawn the figure that cheepychappy adviced for $r_1 \neq r_2$ case. I believe that it is correct strategy as shown in figure-3 to find minimum area among them. Now I need to create equations and solve the problem. enter image description here

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Are there any conditions on the ellipse? If not, you can make it arbitrarily small and the area goes to zero as well. –  Rahul Sep 24 '12 at 10:30
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@RahulNarain : Please consider $a,b,r_1,r_2$ given values and we try to find minimum area among 2 circles and one ellipse. Thanks –  Mathlover Sep 24 '12 at 10:45

2 Answers 2

up vote 2 down vote accepted
+100

Not sure if this helps any, but if you've solved for the $r_1=r_2$ case, I think you should be able to follow from this line of reasoning for the general case.

There exists a triangle with vertices at the centers of the two circles and a third vertex elsewhere. The side which joins the centers of the two circles is of length $r_1+r_2$ and the other two sides contain the radii of each circle pointed towards their respective intersections with the ellipse.

First, optimize the two base angles of this triangle (the angles at the center of each circle, let's call them $\theta_1$ and $\theta_2$) so that the tangents to the circles at each of the three intersections become altitudes of the triangle and so that their intersection is thus the orthocenter of the triangle. Obviously the ratio between $r_1$ and $r_2$ will limit what angles $\theta_1$ and $\theta_2$ can be allowed.

Next, tilt your ellipse so that its major axis is also pointed towards this orthocenter, namely so that its most convex point will overlap the triangle, thus reducing the greatest area possible.

The distances from each tangent point to the orthocenter are equal and each line from the orthocenter to the centers of the circles bisects their angles $\theta_1$ and $\theta_2$.

The segment that joins the two points of intersection of the ellipse is a chord perpendicular to and bisected by the major axis of the ellipse. The optimized solution will have a minimum length for this chord. The angle of elevation of this chord (and thus the minor axis of the ellipse itself) relative to the line connecting the two circles is given by:

$\alpha = \tan^{-1}(\frac{r_2\sin(\theta_2)-r_1\sin(\theta_1)}{r_2(1-\cos(\theta_2)) + r_1(1-\cos(\theta_1))})$

From there, knowing the tilt of the ellipse relative to the line connecting the centers of the two circles and knowing that the angle covering the ellipse's segment is bisected by the major axis, I think you should be able to proceed similarly to the case $r_1 = r_2$. Or at least that's my attempt at an answer before the bounty runs out. Best of luck.

EDIT:

Definitions:

$\theta_1$ is $m\angle CO_2B$

$\theta_2$ is $m\angle AO_1B$,

$\theta_3$ is $m\angle AOC$

(in general, subscript 3 will refer to the ellipse, even though it is not marked as such on the diagram)

Relations between angles:

$\theta_2 = 2\arctan(\frac{r_1}{r_2}\tan(\frac{\theta_1}{2}))$

$\theta_3 = 2\arcsin(\frac{r_1}{b}\tan(\frac{\theta_1}{2})\sin(\frac{\theta_1 + \theta_2}{2}))$

Areas of the three segments:

$A_{\bigcirc_1} = \frac{\theta_1}{2}r_1^2$

$A_{\bigcirc_2} = \frac{\theta_2}{2}r_2^2$

$A_{\bigcirc_3} = a\cdot b\cdot \arctan(\frac{b}{a}\tan(\frac{\theta_3}{2}))$

see this post on finding the area of an ellipse segment

Areas of the triangles that need to be subtracted from those segments:

$A_{\triangle_1} = \frac{1}{2}r_1^2\sin(\theta_1)$

$A_{\triangle_2} = \frac{1}{2}r_2^2\sin(\theta_2)$

$A_{\triangle_3} = \frac{1}{2}(b^2\sin^2(\frac{\theta_3}{2}) + a^2\cos^2(\frac{\theta_3}{2}))\cdot\sin(\theta_3)$

You can use Heron's formula to calculate the area of $\triangle ABC$ using side-lengths defined by

$\overline{AB} = 2r_1\sin(\frac{\theta_1}{1})$

$\overline{BC} = 2r_2\sin(\frac{\theta_2}{2})$

$\overline{AC} = 2b\cdot\sin(\frac{\theta_3}{2})$

The constraint on tangents:

So everything above gives you results only in terms of your constants and the angles, which can all be reduced to being relations of a single angle of your choice. What you need to do to get a solution, is to remember the constraint that the tangent lines of your ellipse need to be the same as your tangent lines to your circle so that the ellipse never overlaps them.

The slope of the tangent lines to your ellipse is given by:

$s_{ellipse} = \pm\frac{b}{a}\cot(\frac{a}{b}\arctan(\frac{\theta_3}{2}))$

As for the circles, you can think of $\alpha$ as defined in my original answer as the circles' rotation away from the $y$-axis, which means that for circle 1 the measure of the angle away from the positive $x$ direction is:

$\theta_1^\prime = -(\theta_1 + \frac{\pi}{2} - \alpha)$

And for circle 2 we get:

$\theta_2^\prime = \theta_2 + \frac{\pi}{2} + \alpha$

Your slopes for these tangents are then given by:

$s_1 = -\cot(\theta_1^\prime)$

$s_2 = -\cot(\theta_2^\prime)$

Setting those slopes equal to the relevant (positive or negative) version of the ellipse's slope and rewriting your angles in terms of a single angle using the relationships given right at the beginning of these edits, should give you a solution for your angle, which then becomes a solution for all three angles, when then should give you a solution for your area... hopefully I haven't made any typos.

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Please let me know your ideas about the Figure-3. Thanks for help. –  Mathlover Oct 1 '12 at 13:12
    
The figure looks great! Here are some equations to relate all the angles. –  cheepychappy Oct 1 '12 at 18:18
    
Ack... can't do linebreaks in comments and it gets messy, so I'll post as separate answer. –  cheepychappy Oct 1 '12 at 18:36
    
Or edit the original... happy solving! –  cheepychappy Oct 1 '12 at 19:13
    
your work is wonderful and really good effort. Thanks a lot for your help. –  Mathlover Oct 1 '12 at 19:56

I don't think I'd have the patience to work through the details, but here's how I'd set things up ...

Parameterize the origin-centered ellipse by $P_\theta := (a\cos\theta,b\sin\theta)$. The tangent vector at $P_\theta$ is $(-a\sin\theta,b\cos\theta)$, so that the outward-pointing normal at $P_\theta$ is $\mathbf{n}_\theta := (b\cos\theta,a\sin\theta)$.

Let $R$ and $S$ be the centers of the circles ---of radius $r$ and $s$, respectively--- and let $$R^\prime := P_\theta \qquad S^\prime := P_\phi$$ be their respective points of tangency with the ellipse (with some appropriate limits on $\theta$ and $\phi$); let $T^\prime$ be the circles' point of tangency with each other.

We have $$R = P_\theta + r \frac{\mathbf{n}_\theta}{|\mathbf{n}_\theta|} \qquad S = P_\phi + s \frac{\mathbf{n}_\phi}{|\mathbf{n}_\phi|} \qquad |R-S|=r+s$$

We can describe the target region as $\triangle ORS$ with some sub-triangles and sectors (two circular, one elliptical) removed. Thus, $$\begin{align} 2A &= 2\left(\triangle ORS-\triangle ORR^\prime - \triangle OSS^\prime-\mathrm{csect} RR^\prime T^\prime-\mathrm{csect} SS^\prime T^\prime - \mathrm{esect}OR^\prime S^\prime \right) \\ &= |R\times S| - |R\times R^\prime| - |S\times S^\prime| - r^2 \;\mathrm{acos} \frac{R\cdot(R-S)}{|R|(r+s)} - s^2\;\mathrm{acos}\frac{S\cdot(S-R)}{|S|(r+s)} - a b \;|\theta-\phi|\end{align}$$ where a little care is taken in properly computing $|\theta-\phi|$ modulo $2\pi$.

From here ... Write $R$ and $S$ in terms of $\theta$ and $\phi$, apply the constraint $|R-S|=r+s$, and go to town.

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