The circle does touch the triangle. Else, it can't be solved.
Lets call the point of intersection (for the one above ) be P and the one below, Q, and the one by the side, R.
AP=AR=x and RB=BQ=y
CP=13-x and CQ=12-y
x+y=5
13-x=12-y
Solving the simultaneous equation yields x=3, y=2.
angle ACB=22.62
angle COP=11.3
tan 11.3 = radius / 10
radius =2