x^2+y^2-10x-10y=0
(x-5)^2+(y-5)^2=50=(5√2)^2
设所求圆方程为(x-a)^2+(y-b)^2=r^2
圆(x-5)^2+(y-5)^2=50=(5√2)^2与(x-a)^2+(y-b)^2=r^2外切
√[(a-5)^2+(b-5)^2]=r+5√2 ①
圆(x-a)^2+(y-b)^2=r^2直线x+√3y=0为半径r
r=|a+√3b|/√2 ②
点M(3,-√3)在圆(x-a)^2+(y-b)^2=r^2上
(3-a)^2+(-√3-b)^2=r^2 ③
解①②③即可求得a,b,r的值!