求与圆x^2+y^2-2x=0内切且与直线 x+√3y=0相切于点M(1,-√3/3)的圆的方程

设方程(x-a)^2+(y-b)^2=c^2
与圆x^2+y^2-2x=0内切
(x-1)^2+y^2=1
(a-1)^2+b^2=(1-c)^2.........................1
与直线 x+√3y=0相切于点M(1,-√3/3)
(a+√3b)/2=c................................2
(1-a)^2+(-√3/3-b)^2=c^2....................3
解得
a=4/3 b=0 c=2/3
圆方程(x-4/3)^2+y^2=4/9

(x-a)^2+(y-b)^2=c^2
与圆x^2+y^2-2x=0
内切
(x-1)^2+y^2=1
(a-1)^2+b^2=(1-c)^2
直线 x+√3y=0相切于点M(1,-√3/3)
(a+√3b)/2=c
(1-a)^2+(-√3/3-b)^2=c^2
a=4/3 b=0 c=2/3
圆方程(x-4/3)^2+y^2=4/9