Question 1094135
<pre>
Let the radius be r,

Then the x-coordinate of the center is the length of the
radius from the y-axis, so it is r.

The y-coordinate of the center is r units below the line
y-1 = 0, which is the same as y = 1, a horizontal line
1 unit above and parallel to the x-axis, in red.

The y-coordinate of the center is 5 units below the red line.
Since the red line is 1 unit above the x-axis, we need to
subtract the radius from 1 to get the y-coordinate, That's
1-r, so that's the y-coordinate of the center, and the 
center is (r,1-r).   

{{{drawing(400,400,-3,11,-10,4,
blue(line(-5,1,12,1)), green(line(0,-4,5,-4),line(5,1,5,-4)),
graph(400,400,-3,11,-10,4),circle(1,0,.1),circle(2,0,.1),
locate(5,-4,"(r,1-r)"),circle(5,-4,.1),
circle(1,0,1), circle(5,-4,5) )}}} 

The center is (r,r-1) and the radius is r
so the equation of the circle, 

{{{(x-h)^2+(y-k)^2=r^2}}}

becomes:

{{{(x-r^"")^2+(y-(r-1)^"")^2=r^2}}}

and since it goes through (2,0)

{{{(2-r^"")^2+(0-(r-1)^"")^2=r^2}}}

{{{(2-r^"")^2+(-r+1)^"")^2=r^2}}}

{{{4-2r+r^2+r^2-2r+1=r^2}}}

{{{2r^2-4r+5=r^2}}}

{{{r^2-4r+5=0}}}

{{{(r-1)(r-5)=0}}}

r-1 = 0;   r-5 = 0
  r = 1      r = 5

The small circle has r=1, center (r,1-r) = (1,1-1) = (1,0)

{{{(x-1^"")^2+(y^"")^2=1}}}

and the large circle has r=5, center (r,1-r) = (5,1-5) = (5,-4)

{{{(x-5^"")^2+(y+4^"")^2=25}}}

Edwin</pre>