The Praying Eyes theorem generalized Is there an obvious way to slice two spheres (no necessarily equal) 
simultaneously such that the sections share common areas? I mean, I cannot see another way (easier) different from the way provided by the Praying Eyes theorem and related theorems.
You may want to take a look of this first:
http://geometriadominicana.blogspot.com/2014/03/praying-eyes-theorem.html
Edit: just wanted to say that the construction is possible for three spheres too. For the construction in 2D take a look of this page http://www.cut-the-knot.org/pythagoras/LightingTheBall.shtml

 A: EDIT: the construction far below suffices when the center P is outside the circle with center Q, including when the circles do not intersect at all. For the missing case, I'm thinking a segment through an intersection point of the two circles, the only requirement being same chord length. I'll think about it. 
EDIT TOO: if the circles intersect, we can simply take their intersection as the collinear chords of equal length: 

ORIGINAL:This is actually a two dimensional construction, no continuity argument required; also, it is sometimes impossible. 
Start with a circle with center P of small diameter (red) and a circle of center Q of larger diameter. Draw the red segment as a chord of the second circle, and find the segment midpoint W. Draw another circle around center Q, this time with radius QW. 
Next, draw the midpoint of the segment PQ, at M. Draw a semicircle around M with radius MQ. This intersects the QW circle in a point, call this T. 
By construction, QTP is a right angle. Since length QT is the same as QW, the intersection of the line PT with the larger circle is the same length as the diameter of the smaller circle, always in red. 
In higher dimension, instead of just the line PT, take the codimension one hyperplane... that is, take any 2-plane through both centers and perform this construction, then fill in

