答:
连接BC,取中点D((x1+x2)/2,(y1+y2)/2]
AD^2=[x0-(x1+x2)/2]^2+[y0-(y1+y2)/2]^2
BD^2=[x1-(x1+x2)/2]^2+[y1-(y1+y2)/2]^2
R^2=AB^2=AD^2+BD^2
R^2=[x0-(x1+x2)/2]^2+[y0-(y1+y2)/2]^2 + [x1-(x1+x2)/2]^2+[y1-(y1+y2)/2]^2
答:
连接BC,取中点D((x1+x2)/2,(y1+y2)/2]
AD^2=[x0-(x1+x2)/2]^2+[y0-(y1+y2)/2]^2
BD^2=[x1-(x1+x2)/2]^2+[y1-(y1+y2)/2]^2
R^2=AB^2=AD^2+BD^2
R^2=[x0-(x1+x2)/2]^2+[y0-(y1+y2)/2]^2 + [x1-(x1+x2)/2]^2+[y1-(y1+y2)/2]^2