The center of the inscribed circle of a triangle is the intersection of three bisectors.

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.