To solve this problem, use the facts that the center of a internal circle, which is inscribed in a triangle, is equidistant from the 3 sides of triangle, and the center of a external circle, which circumscribes a triangle, is equidistant from the 3 vertices of triangle.
First, we draw the triangle. Draw a horizontal ray, Mark point A and B on the ray. Point B is 4cm away from A to the right. Draw 2 arcs crossing above AB: one from A with radius 6cm, the other from B with radius 5cm, and label the crossing of 2 arcs C. So, AB=4cm, BC=5cm and CA=6cm.
Second, we draw the internal circle. Bisect ∠A and ∠B, and label the crossing of 2 angle bisectors above AB, O. From O, draw 3 perpendiculars to AB, BC and CA, crossing AB at G, BC at H and CA at I. Draw a circle around the center O with radius OG. Circle O is the required internal circle that only touches the triangle at points G, H and I. Because △AOG≡△AOI, △BOG≡△BOH and △COH≡△COI, so OG=OH=OI.
Third, we draw the external circle. Draw 2 perpendicular bisectors of segment AB and BC, and label the crossing of 2 bisectors above AB, P. Draw a circle around the center P with radius PA. The circle P is the required external circle that passes thru 3 points, A, B and C. Because △PAB,△PBC and △PCA are isoscelese triangle, so PA=PB=PC.
Therfore, circle O is the required internal circle with radius approx. 1.32cm, and circle P is the required external circle with radius approx. 3.02cm.