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...