Circle And Circle In A Closed Curve

 Consider,

where \(a'\) and \(b'\) are two points along the perpendicular bisector of  the diameter \(ab\), intersecting with the arbitrary closed curve.   A blue circle is drawn using \(a'b'\) as diameter.  When the lengths \(ab=a'b'\) and the red center of  the circle defined by diameter \(ab\) will then coincide with the blue center of the circle define by diameter \(a'b'\), by adjusting point \(a\) and \(b\), we have one circle along the circumference which points \(aa'bb'\) forms a square.

Both red center and blue center will move as \(a\) and \(b\) are moved.

The blue center will move erratically, as \(a'\) and \(b'\) hop along the closed curve.  This is because line \(ab\) changes length (thus a new center) and rotates as points \(a\) and \(b\) move.  The red center will trace out a smooth locus as \(a\) and \(b\) move along the continuous curve. 

Does this method of drawing a square in the closed curve guarantee a solution?  That a square is possible always?

Move the red center towards the blue center always, util both centers meet and \(ab=a'b'\).

Plenty of fun moving all over the closed curve.

Note:  In this example, as the diagram indicates, move point \(b\) upwards rotates the perpendicular bisector \(a'b'\) and may result in \(ab=a'b'\).

Or move point \(b\) downwards to rotate the perpendicular bisector \(a'b'\) clockwise may result in \(ab=a'b'\).