Circles tangent to opposite sides have as diameter the width and length of the rectangle.
Hence can be equal only if the rectangle is a square.
There are then infinitely many solutions.
Point I is dragable/animated in the applet to chose a pair of circles.
The "rectangle" ABCD aspect ratio is defined by the dragable opposite corners A and C (initially a square)
We then consider only circles, each tangent to adjacent sides !
There are several solutions, depending on the circles being "inside" or "outside" the rectangle.
At symmetries, there are 4 families of circles, depending on the respective angles which "contain" them.
Without loss of generality, we may consider the opposite angles A and C.
Both angles outside is impossible : the circles couldn't touch !
Angle C is then an inside angle, and the 4 choices for angle A give the 4 families.
By symmetries, this results into xx solution pairs of circles.
The problem is to construct these circles, at least one "representant" for each family.
An applet showing a representant in each family.
The rectangle is defined by its opposite dragable corners A and C.
The other circles in a family are not shown, but result from symmetries and are then circles equal to this representant.
We may note that the first family has two different representants : red and magenta.
The red circles are the only case when both circles are inside the rectangle, if the aspect ratio of the rectangle is in 1/2 ≤ BC/AC ≤ 2