Quarter-circle

Let O be the center of the quarter-circle

Let M    be the intersection point of the edges of  both circles

Draw  a tangent line  to  the quarter circle at M and  let N be any other point on this line

Angle  OMN = 90°

Let P be the center of the small circle

Angle PMN  also   = 90°   and PM   must be the radius of the smaller circle since it forms a 90° with the tangent line

Then  PM  must lie  on  OM  since angle PMN  = angle OMN

Let S be the other intersection of the radius of the quarter circle and  the eadge of the  smaller circle

PS =  radius of the small circle = r

OP  =  OM - PN  =   4 - r

OS = 2 + r

Triangle  OPS  forms a right triangle such that

OP^2  =  PS^2  + OS^2

(4 - r)^2  =  r^2  + ( 2 + r)^2

r^2 - 8r + 16 =  r^2  + r^2 + 4r + 4

r^2  + 12r - 12  =  0

r^2 + 12r  =  12                complete the square on r

r^2 + 12r + 36   = 12 + 36

(r + 6)^2  =  48         take the positive root

r + 6   = sqrt (48)

r = sqrt (48)  -  6

r = 4sqrt (3)  - 6    ≈  .923