euclidean-geometrygeometric-constructionplane-geometry
We can use the compass and straightedge to find the center of one circle.
We have proven we cannot find the center of a circle with straightedge alone (see: http://www.cut-the-knot.org/impossible/straightedge.shtml)
But if we are given two intersecting circles, can we find the centers of both circles?
All of these constructions are fairly complex, and I'm not going to give them in any detail. OP has asked for information that will help move the ball forward. I'll give some comments and references, and will add detail or clarification or corrections where requested. I have not vetted any of the constructions in every detail. I'll number the claims to match the numbering in the question.
Two non-intersecting circles with a point on the join of the centers. The site referred to by OP is most likely http://mathafou.free.fr/pbg_en/sol100a.html. The construction is the topic of C. Gram. A remark on the construction of the centre of a circle by means of the ruler. Math. Scand., 4:157–160, 1956.
Three non-intersecting circles. Apparently in D. Cauer. Uber die Konstruktion des Mittelpunktes eines Kreises mit dem Lineal allein (Berichtigung). Math. Ann., 73(1):90–94, 1912 and 74(3):462–464, 1913 (German). If you can find Smogorzhevskii's The ruler in geometrical constructions, the construction for three circles not in the same coaxial system is given at the end of the book.
Two congruent non-intersecting circles with an arbitrary point on their radical axis. Use the following construction to get the line $X_{11}X_{12}$, which joins the circle center. After that use the same construction as in 3. In the figure, givens are in blue, and the points are numbered in order of creation. The four chords (e.g. $X_1X_2$) are polars of $P$ and X_6.
Any arc of the circle with the center identified. See Chris Impens' Geometry without compasses
Bonus case. Two non-congruent non-intersecting circles with an arbitrary point on their radical axis. As for case 5, build the points $X_1,\dots,X_5$. Let $X_E$ be the intersection of the lines $X_2X_3$ and $X_1X_4$. Then line $X_5X_E$ goes through the centers.
Akopyan and Fedorov's Two circles and only a straightedge gives a useful overview of some of these topics.
If you place one end of the compass at A and the other at B, you are allowed to draw either a circle centered at A or a circle centered at B - nothing else. You may be thinking that you can lift the compass off the plane and transfer it to another point while retaining the separation of the two points of the compass by the distance d. In classic straightedge-and-compass constructions, this is explicitly not allowed - the compass is considered to be 'collapsible', i.e the joint where the two arms of the compass meet is 'loose', so the moment you lift the compass, the plane is no longer able to hold the two arms at distance d.
In the case at hand: you can draw a line bisecting the angle between the X and Y axes, and mark the point P' where it intersects the circle about the origin of radius $\sqrt2$. It is not hard to see that this point is at a distance 1 from the X and Y axes, so you can find a point of distance 1 on (say) the Y axis by drawing a line parallel to the X axis through P'. You can then draw a circle of radius 1 centered at P', intersecting the first line at P''. A circle centered at the origin through P'' will then intersect the X axis at a distance $1 + \sqrt2$ from the origin.
Best Answer
For a chord $HG$ of the first circle, we can construct two parallel chords $EF$ and $IJ$ of the second circle. Since $EFIJ$ is an inscribed trapezoid, $LK$ is a perpendicular bisector of its base and passes through a circle centre. Two of those give you the centre.