取BC边中点D并联结AD.
∵AB=AC
∴△ABC为等腰三角形
∵D为BC中点
∴AD⊥BC且BD=1
则AD=√(9-1)=2√2
则S△ABC=1/2×AD×BC=2√2
设内切圆半径为r,内接圆圆心为o.
则S△ABC=S△oAB+S△oBC+S△oAC=1/2×r×C(C为周长)=1/2×r×8=4r=2√2
则r=√2/2.
设外接圆半径为R,外接圆圆心为O.联结OA,OB,OC,OD
则有OD²+BD²=OB²
(AD-AO)²+BD²+OB²
(2√2-R)²+1=R²
8-4√2R+1=0
4√2R=9
R=9/(4√2)=9√2/8