过原点,且直线x=1及圆(x-1)^2+(y-2)^2=1相切的圆的方程.

设：圆的方程为(x-a)^2+(y-b)^2=r^2
∵过原点,∴a^2+b^2=r^2
∵ 与直线x=1相切,∴│a-1│=r
∵ 与圆（x-1）+(y-2)^2=1相切,∴√（a-1)^2+(b-2)^2=1+r
解得:a=3/8 ,b=1/2 ,r=5/8
圆的方程：（x-3/8)^2+(y-1/2)^2=(5/8)^2