We plot the line 2x-y=3 by finding its y-intercept (0,-3), and we
know that it must go through (2,1), and we sketch the circle approximately: 


 
 The circle has the equation:
 
 (x-h)²+(y-k)² = r²
 
with center (h,k) and radius r, but since the center is on the 
x-axis, we know that k=0, so its equation is:
 
(x-h)²+y² = r²
 
We know that a radius drawn to the point of tangency is 
perpendicular to a tangent line, so we find the equation 
of the line perpendicular to the give line at the point
of tangency, (2,1).  The green line drawn below:
 


First we have to find the slope of the given line

2x - y = 3

We put it in slope-intercept form to find the slope:

2x - y = 3
    -y = -2x + 3
     y = 2x - 3

Compare that to 
  
     y = mx + b

And we see that the slope of the given tangent line is 2
Therefore the slope of the green line has a slope which is
the "negative reciprocal" of 2, which is .  Now
we use the point-slope form to find the equation of the
green line:

y - y1 = m(x - x1)
y - 1 = (x - 2)
y - 1 = x + 1 
    y = x + 2

That's the equation of the green line, so if we find its
x-intercept, that will be the center of the circle. So
we set y=0 to find the x-intercept:

    0 = x + 2
Multiply through by 2

    0 = -x + 4
    x = 4
So now we know that the x-coordinate of the center of the
circle is 4 so the center is (h,k) = (4,0) and the equation of
the circle is

(x-4)²+y² = r²

But we still do not know the radius of the circle.  We can
either find the distance from (4,0) to (2,1) using the
distance formula. But there is an easier way.  We can just
substitute the point (2,1) into the equation of the circle:

(x-4)²+y² = r²
(2-4)²+1² = r²
  (-2)²+1 = r²
      4+1 = r²
        5 = r²

We can find the radius as , but all we want is
the equation, and we need r² for that. So the answer is:

(x-4)²+y² = r²
(x-4)²+y² = 5

Now we can show the complete graph:



Edwin