圆A:x^2+y^2+4x+2y-13=0,
设圆B的方程为(x-m)^2+(y-3m)^2=r^2,
相减得(4+2m)x+(2+6m)y=13+10m^2-r^2,①
点(-2,-1)在①上,
化简得r^2=10(m+1/2)^2+41/2,
所求圆B的方程为
(x+1/2)^2+(y+3/2)^2=41/2.
圆A:x^2+y^2+4x+2y-13=0,
设圆B的方程为(x-m)^2+(y-3m)^2=r^2,
相减得(4+2m)x+(2+6m)y=13+10m^2-r^2,①
点(-2,-1)在①上,
化简得r^2=10(m+1/2)^2+41/2,
所求圆B的方程为
(x+1/2)^2+(y+3/2)^2=41/2.