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Let $r = a-c$, where $a$ is the length of semi major axis and $c$ is the distance between origin and one of the foci of an ellipse.

I'm wondering if there a nice way to see that the below circle is always interior to the ellipse. $$(x-c)^2 + y^2 = r^2$$

I think this is equivalent to saying that the minimum distance between a focus and a point on the ellipse is $a-c$. If so, is there a way to prove this using geometry or some other means ? I feel calculus is messy here... Thank you!

https://www.desmos.com/calculator/il5rpye8on

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If the semi-minor axis is $b$, then $c^2 = a^2-b^2$.

For the ellipse, $(x/a)^2+(y/b)^2 = 1$ or $y^2 =b^2(1-(x/a)^2) =b^2-(bx/a)^2 $.

For the circle,

$\begin{array}\\ y^2 &=r^2-(x-c)^2\\ &=(a-c)^2-(x-c)^2\\ &=a^2-2ac+c^2-(x^2-2xc+c^2)\\ &=a^2-2ac-x^2+2xc\\ \end{array} $

So, we want to show that $a^2-2ac-x^2+2xc \le b^2-(bx/a)^2 $ or $x^2(1-(b/a)^2)-2xc \ge a^2-2ac-b^2 $ or $x^2(c/a)^2-2xc \ge c^2-2ac $ or $x^2(c/a)^2-2xc+a^2 \ge c^2-2ac+a^2 $ or $(cx/a-a)^2 \ge (c-a)^2 $.

Since $a \ge c$ and $x \le a$, $cx/x \le a$, so this last is equivalent to $a-cx/a \ge a-c$ or $cx/a \le c$ or $x \le a$, which is true.

Therefore the circle is always within the ellipse.

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Think to the gardner-method to construct the ellipse: then is clear that your circle is the curvature circle at the perihelion (aphelion), and thus internal tangent to the ellipse (by Newton/Kepler laws).

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    $\begingroup$ Internal tangency doesn't guarantee that there's no point of intersection somewhere else, though, without further arguments. $\endgroup$ – Steven Stadnicki Mar 7 '17 at 1:21
  • $\begingroup$ @StevenStadnicki: as a quick argument, I proposed a Mechanical argument: the perihelion is by definition the nearest point to the focus=sun, so the curvature circle there (=tangent circle centered at focus) cannot intersect the ellipse in other points. $\endgroup$ – G Cab Mar 7 '17 at 9:32
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This is fairly easy to prove using polar coordinates. Move the origin to the focus at $(c,0)$ and rotate so that $\theta=0$ points in the direction of the ellipse’s center. The equation of the ellipse is then $\rho={a(1-e^2)\over1+e\cos\theta}$, where $e$ is the eccentricity of the ellipse. The eccentricity is equal to $c/a$, and substituting this into the above equation produces $$\rho={a(1-(c/a)^2)\over1+(c/a)\cos\theta} = {a^2-c^2\over a+c\cos\theta} \ge {a^2-c^2\over a+c}=a-c.$$ (The denominator on the left side of the inequality is always positive since $a>c$.).

If you want to get fancy, you could construct an argument using the curvature of the ellipse at and away from its intersection with the circle.

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