Mfcttrf

I have two two incircle between rectangle and two
quadrilateral circlein. It's possible to determine exact value of $phi,$ angles between vectors of center of two circles.

Place the origin $O$ at the lower left-hand corner, and let the positive $x,y$-axes be the the natural choices based on the diagram.

Let $r$ be the radius of the red circle.

Let $P,Q$ be the centers of the circles of radii $2,1$, respectively.

Then we can express the coordinates of $P,Q$ as $P=(x,2)$ and $Q=(x,r-1)$, where $x$ is an unknown.

Computing $|OQ|$ in two ways, we get the equation
$$sqrtx^2+(r-1)^2=r+1$$
hence
$$x^2+(r-1)^2=(r+1)^2tageq1$$
Computing $|OP|$ in two ways, we get the equation
$$sqrtx^2+2^2=r-2$$
hence
$$x^2+4=(r-2)^2tageq2$$
From $(texteq1),-,(texteq2)$, we get $r=8$.

Plugging $r=8$ into $(texteq2)$, we get $x=4sqrt2$.

Using the known values of $r$ and $x$, the distance formula yields
beginalign*
|OP|&=6\[4pt]
|OQ|&=9\[4pt]
|PQ|&=5\[4pt]
endalign*
hence by the law of cosines, $cos(largephi)=largefrac2327$, so we get $largephi=cos^-1bigl(largefrac2327bigr)$.

Using Pythagoras theorem is not difficult to find that the radius $B$ of red arcs is $8$. It follows that $PQ=6$, $PR=9$ and $QR=5$. Apply now the cosine rule to triangle $PQR$ to find
$$
cosphi=23over27.
$$

Let $|AB|=|CD|=a$,$|BC|=|AD|=b$,
$|O_1E|=r_1$,
$|O_2F|=r_2$,
$angle O_1AO_2=phi$,
$angle O_1AF=alpha$
$angle O_2AF=beta$.

Then
beginalign
tanalpha&=fracb-r_1a/2
tag1label1
,\
sinalpha&=fracb-r_1b+r_1
,\
tanalpha&=fracsinalphasqrt1-sin^2alpha
=fracb-r_1b+r_1
left/
sqrt1-Big( fracb-r_1b+r_1 Big)^2 right.
=tfrac12,fracb-r_1sqrtb,r_1
tag2label2
.
endalign

From eqref1$=$eqref2 it follows
beginalign
a&=4,sqrtb,r_1
tag3label3
.
endalign

Similarly,
beginalign
tanbeta&=frac2r_2a
tag4label4
,\
sinbeta&=fracr_2b-r_2
,\
tanbeta&=
fracsinbetasqrt1-sin^2beta
=fracr_2sqrtb,(b-2,r_2)
tag5label5
.
endalign

From eqref4$=$eqref5:
beginalign
a&=2,sqrtb,(b-2,r_2)
tag6label6
,
endalign

and from eqref3$=$eqref6
we have

beginalign
b&=4r_1+2r_2
,\
a&=4,sqrtr_1,(4,r_1+2,r_2)
.
endalign

beginalign
tanphi&=tan(alpha-beta)
=fractanalpha-tanbeta1+tanalphatanbeta
=frac2(3r_1+r_2),sqrt2,r_1,(2,r_1+r_2)16,r_1^2+11,r_1,r_2+2,r_2^2
endalign

For $r_1=1$, $r_2=2$ we have

beginalign
a&=8,sqrt2approx 11.31370850
,\
b&=8
,\
phi&=arctanBig(frac10,sqrt223 Big)
approx 31.586338^circ
.
endalign

There are constraints that $A$ and $B$ must fullfil, i.e., the following system of equations coming from Pythagoras theorem applied to certain right triangles :

$$begincases(B-2)^2+2^2=(A/2)^2\(B-1)^2+(A/2)^2=(B+1)^2endcases$$

giving $A=8 sqrt2$ and $B=8$.

If now we take equations as in the partial solution you gave (a good idea) :

$$tan(theta) = dfrac2A/2 = dfracsqrt24tag1$$

and

$$tan(phi + theta) = dfrac74 sqrt2tag2$$

Equation (2) can also be written :

$$dfractan(phi) + tan(theta)1-tan(phi)tan(theta) = dfrac7sqrt28tag3$$

from which we can extract the numerical value of $tan(phi)$. Up to you...