Let the tangent AC side be e and the tangent BC side be f.
According to the properties of angular bisector: AD=AE, CE = CF
BD=BF
Then |CA|-|CB|=|AE|+|CE|-(|CF|+|BF|)
=|AD| + |CF| - |CF| - |BD|
=|AD| - |BD|
The center of the circle moves on the straight line x=2.
∴|OD|=2
Then |AD|=√5+2, |BD|=√5-2.
I think it will be in the back.