Hi MATNTRNG,
One way is to show that if we draw 2 more circles,
same radii as Circle1 and Circle 2, but centred on the original circle's centres as shown,
the geometry is entirely symmetrical since triangle FmG is isosceles,
due to the two chords being of equal length.
We can repeat the geometry under the centreline,
hence point "m" is on the centreline,
and due to the left-right symmetry, also must be midway between the circle centres.
The previous post is really not suitable as a proof, MATNTRNG.
Instead, the attachment shows that if the chords are to be of equal length,
then they form the basis of numerous isosceles triangles, two of which are shown.
Since the chords themselves form isosceles triangles with the circle centres, then
$\displaystyle [m_1o_1]$ and $\displaystyle [m_2o_2]$ are parallel.
These line segments are also parallel to the perpendicular line passing through
the point at which the chords meet.
The point on the joint-centreline of the two circles that is equidistant from F and G lies on that perpendicular line
and therefore is midway between the centres.