Find the radius of the circle passing through the point (2,2) and having x-y-3=0 as its diameter.

It is possible to draw a circle with diameter x - y - 3 =
0 passing through (2,2) with any radius, but there is a minimum radius that can be drawn
when the line through the center of the circle is perpendicular to (2,2).   This
perpendicular line will have a slope that is the negative reciprocal of the diameter. 
The slope of the diameter is 1 so the slope of the perpendicular line would be -1/1 =
-1.

A line through (2,2) with slope -1 is y - 2 = -(x - 2) 
or y = -x + 4

y = -x + 4 intercects x - y - 3 = 0 (or y = x
- 3) when -x + 4 = x - 3

solving for x we get 2x = 7 or x =
7/2 substituting we get y = 1/2, so the point is at (7/2, 1/2) which you can verify is
the only point which is on y = -x + 4 and y = x - 3.

The
radius is the distance between the center (7/2, 1/2) and
(2,2).

Distance = sqrt((7/2-2)^2 + (1/2 - 2)^2) =
sqrt((3/2)^2 + (-3/2)^2) = sqrt(9/4 + 9/4)

Which gives 3/2
sqrt(2) as the minimum radius.  Moving the center left or right of this point increases
the radius so the minimum radius of a circle with diameter x - y - 3 = 0 and (2,2) a
point on the circle is 3/2 sqrt(2).