The locus of the points equidistant from two given points is the perpendicular bisector to the segment connecting these points.

So, you need to construct the perpendicular bisector to the segment connecting the given points 
P = (7,-2) and Q = (5,0), and then find its intersection with the given straight line.



The midpoint between the two given points is (6,-1).

The segment connecting (7,-2) and (5,0) has the slope  =  = -1.

Hence, the perpendicular line (perpendicular bisector) has the slope 1.

The line with the slope 1 passing through the point (6,-1) has the equation 

y - (-1) = (1)*(x-6),   or

y = x - 7.

The intersection of the straight lines

x - 2y - 9 = 0     (1)   (the given line)   and
y = x - 7          (2)   (the perpendicular bisector)

is (solve the system by substitution) the point (x,y) = (5,-2)

So, the center of the circle is the point (5,-2).

 
  


        

  
  
The radius of the circle is the distance from the point (5,-2) to the point (5,0) is 2 units

Hence, the radius of the circle is 2.


Therefore, the equation of this circle is 

 = 4.