Label each vertex, (0,0), (4,0) & (0,3), A, B & C respectively. Draw △ABC on a graph. ∠A=90°, △ABC is the 3:4:5 right triangle, and BC=5. Draw an imaginary inscribed circle O of its radius r, and label the three points of contact with sides, AB,BC & CA, L, M & N respectively. So OL=OM=ON=r, and OL⊥AB, OM⊥BC & ON⊥CA. Therefore △OAL≡OAN, △OBL≡△OBM & △OCM≡△OCN (RHS). (So OA, OB & OC bisect ∠A, ∠B & ∠C respectively.)
As defined above, ∠A=∠OLA=∠ONA=∠LON(360°－270°)=90°, and OL=ON=r. So quadrilateral ALON is a square, and the coordinates of O are x=AL=r & y=AN=r, that is O(r,r).
BM=BL=AB－AL=4－r, and CM=CN=CA－AN=3－r, and BC=5 ⇒ BC=BM+CM=(4－r)+(3－r)=5 ⇔ r=1 Therfore the coordinates of the center of inscribed circle, O, are O(1,1).
On the other hand, the center of circumscribed circle is found using the fact that the angle subtended at the circumference by the diameter is a right angle. The converse is also true. Therefore BC is the diameter of circumscribed circle and the midpoint of BC is the center: the coodinates are (2,1.5).