How to prove the path has minimal length?

Question

Setting：Given an isosceles right triangle AEC，A=90 degree，AE=AC. Given r>0，we draw circle A and circle E with radius r. Let point G=the midpoint of the "quarter circle A". We draw a line L parallel to segment CG and the distance of CG and L is r. Let point M=the intersection of L and segment CE. We draw circle M with radius r.

Goal：We want to find point O on circle M and point P on circle E such that the sum GO+OP or GP+PO has minimal length.

Conjecture：(1)Clearly when r is large such that segment GE and circle M has intersection，then O、P are the intersection of GE and circle M、E，respectively. (see picture 1)

(2) when r is small such that segment GE and circle M has "NO" intersection，then the minimal sum is GO+OP (Not GP+PO)，where points O、P satisfy： (a)An ellipse with two foci G、P is tangent to circle M at tangent point O (b)segment OE and circle E intersect at P (see picture 2)

p.s.：The hard part is the order of path can be G->circle M ->circle E or G->circle E->circle M，and we want to find minimal path among the 2 cases. Hence we must prove： Given arbitrary 2 points N on circle M and L on circle E，the inequality GO+OP <GL+LN is true.

I checked the conjecture (2) by software "Geogebra"，it seems right，but I don't how to prove. Please help me，thanks.

Related pictures：

Answer 1

Once you choose $O$ on circle $M$, then the point $P$ on circle $E$ with the least distance from $O$ is the one on segment $OE$. Note that $OG+OP=OG+OE-r$: as $r$ is fixed, $OG+OP$ is minimum when $OG+OE$ is minimum.

Consider now the ellipse through $O$ with foci $G$ and $E$. If it isn't externally tangent to circle $M$, then some point $O'$ on that circle would be within the ellipse and $O'G+O'E<OG+OE$, that is point $O$ would not give the minimum sum of distances from $G$ and $E$. It follows that $O$ must be chosen so that the ellipse is externally tangent to circle $M$: in that case all other points on circle $M$ are external to the ellipse and $OG+OE$ is minimum.

By the way: I've found here a "simple" geometric construction of the ellipse.