Problem: Show that $ O_1O_2 = m + n $, where $ 2 $ m and $ 2 $ n are the axes of the ellipse. ($ r $ and $ s $ are parallel)
I need a big help, so thank you!
Attemp:
I first tried to relate how far a circle and an ellipse are when the circle tangents it internally
Attemp 2:
I am investing in this second attempt: I put a circle tangent to the ellipse on the outside and trying to find the relationships that exist inside and out.
I also tried to mess with cylinders. I tried to put the ellipse inside the cylinder to mess with some lemmas
The centres $O$ and $C$ of circle and ellipse must lie on the midline equidistant from $r$ and $s$. We can choose at will a point $P$ on the circle ($\theta=\angle COP$) and another point $T$ on line $r$: I'll show a construction to obtain an ellipse tangent to $r$ at $T$, to the circle at $P$ and to $s$.
Construct first of all the common tangent at $P$, perpendicular to radius $OP$, and let it intersect line $r$ at $Q$. If $M$ is the midpoint of $PT$, then the centre $C$ of the ellipse is the intersection between line $QM$ and the midline.
$TC$ is a semi-diameter of the ellipse, having its conjugate diameter along midline $OC$. To find the conjugate semi-diameter $AC$ we can use the ellipse equation: $$ {PH^2\over AC^2}+{CH^2\over CT^2}=1, $$ where $H$ is the intersection between line $CT$ and the parallel to $r$ passing through $P$. But $CH/CT=\sin\theta$, hence: $$ AC={PH\over\cos\theta}. $$ That completes the construction. If we set: $r=OP$, $x_0=DT$, then it is not difficult to find that: $$ AC=x_0{1+\sin\theta\over\cos\theta}-r, \quad CT^2=x_0^2{\sin^2\theta\over(1-\sin\theta)^2}+r^2, \quad OC={x_0\over1-\sin\theta}. $$
To prove the requested equality, we can use Apollonius' identities, relating semi-axes $m$ and $n$ to conjugate semi-diameters $AC$ and $CT$:
$$ mn=AC\cdot CT\cdot\sin(\angle ACT)=AC\cdot OP, \qquad m^2+n^2=AC^2+CT^2, $$ which give: $$ (m+n)^2=AC^2+CT^2+2AC\cdot OP. $$
Inserting here the expressions given above, we can then check that $m+n=OC$, as it was to be proved.