Is this circle always interior to ellipse? 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
 A: 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).
A: 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.
A: 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.
