SOLUTION: Find equation when the center on the line x - 2y - 9 = 0 and passes through the points (7,-2) and (5,0).

The center of the circle lies on the perpendicular bisector to the segment connecting the points (7,-2) and (5,0).

The midpoint is (6,-1).     (6 = ,  -1 = )


The segment connecting the given points has the slope  =  = -1.

Therefore, the perpendicular bisector has the slope 1.

Hence, the perpendicular bisector passing through the point (6,-1) has the equation

y - (-1) = 1*(x-6),   or,  which is the same,  y+1 = x - 6.


Hence, the center of the circle lies at the intersection of these two lines

x - 2y - 9 = 0    (1)   and
y+1 = x - 6.      (2)

This system is the same as

x - 2y = 9,       (3)
x -  y = 7.       (4)

To solve the system (and to find the intersection point) subtract (4) from (3) both sides). You will get

-y = 2,   or   y = -2.

Then x = 7 + y = 7 + (-2) = 5   (from (4).

Thus the point (5,-2) is the center.


The distance from the center (5,-2) to the given point (5,0) is  =  =  = 2.


Then the equation of the circle is 

 = 4,   or

 = 4.