Prove that the line, containing the segments with lengths the max and min distances from the origin to a circle, contains the center of the circle (https://www.desmos.com/calculator/nwdvygfw1r for reference)
I have a problem trying to prove what my intuition is telling me.
I was trying to find the maximum and minimum distances from the origin of the 2D plane to a circumference, and I thought that maybe the distance's extrema formed line segments that when lied up together, were contained in the line formed with the origin and the circumference's center.
My strategy then was to find the equation of the line passing through the origin and the circumference center, find the points at which it cut the circumference, and thus calculate the distances from those points to the origin, taking the larger one as the maximum distance and vice versa.
I was stumped when I asked myself why exactly my thoughts led me to blindly believe that those two distances should be contained in that line. When I tried to prove it, I thought of brute-forcing it, by calculating the distance formula ( d(O,P), with P being a point on C ), differentiating it, and then setting it equal to zero, but ended up with a hot mess.
My attempt for a circumference with the origin inside it proved successful, but I was unable to extend the rationale of the triangle inequality to the case where the origin is external to the circumference, much less when it lies on the circumference.
Any help is appreciated! This is my first time posting here and I'm hoping I can get somewhere with this problem.
(in Desmos link: how can I prove that if OA and OB are the min and max distances (A,B points on the ciruference), from O to the circumference respectively, then line AB contains C?)
 A: Let $A$ and $B$ be the endpoints of the diameter passing through the point $O$ such that $OA<OB$.
If $P$ is any point of the circumference such that $P\ne A$ and $P\ne B$, we consider the triangle $OCP$.
By applying the property of triangles that states that each side is longer than the difference of the other two, we get that
$OP>CP-OC$
but
$CP=CA$
because they are radii of the same circumference, so we get that
$OP>CA-OC=OA$
It means that $A$ is the point of the circumference of minimal distance from $O$.
Now we apply to the triangle $OCP$ the property that states that each side is shorter than the sum of the other two,
$OP<CP+OC$
but
$CP=CB$
because they are radii of the same circumference, so we get that
$OP<CB+OC=OB$
It means that $B$ is the point of the circumference of maximal distance from $O$.
So the points of minimal and maximal distance are $A$ and $B$ which are the endpoints of the diameter passing through the point $O$.
Since the segment which connects the points $A$ and $B$ of minimal and maximal distance is a diameter, it contains the centre $C$ of the circumference.
A: We can take the property of a circle radius $a$ that for a variable line the product of segment lengths is a constant. Using polar coordinates origin $C$ to set up equation of an eccentric circle.
Let
$$ OP=r, OG =T, CP =CQ=a, OQ= r -2 a \sin \psi$$
Center $C$ can be anywhere, not necessarily on the x-axis.

$$ OP\cdot OQ= r (r-2a \sin \psi )= T^2 $$
Quadratic equation in $r\;$ has two roots.We take positive sign before radical for segment length. At tangent point
$$ \psi=0,\;r_{tgt}= T= OG $$
At far off  point $F$ is maximum distance
$$  \psi= +\pi/2,\; r^2-2 ar-T^2=0 \rightarrow r_{max}= a+ \sqrt{a^2+T^2}=+OC+a$$
At nearby  point N is  minimum distance
$$ \psi= -\pi/2, \;r^2+2 ar-T^2=0 \rightarrow r_{min}= - a+ \sqrt{a^2+T^2}=+OC-a. $$
Since the average distance $ OC = \sqrt{T^2+a^2}$ is constant, it is proved extrema occur when $NF$ is diameter of circle.
A: Construct a circle about $O$ with radius $OA$ and a circle about $O$ with radius $OB.$

The diameter of each circle lies on the extended line $AB.$ The two smaller circles pass through $A$ and the two larger circles pass through $B.$
Show that the smallest circle is completely contained within the circle about $C$
(except for a point of tangency at $A$) and the circle about $C$ is completely contained within the largest circle (except for a point of tangency at $B$).
Hence all points of the circle about $C$ (other than $A$) are further from $O$ than $A$ is and all points other than $B$ are closer than $B$ is.
