The standard equation of a circle passing through points a (5,2) and B (3, - 2) with the center of the circle on the line 2x-y-3 = 0 I will find out the straight line L through AB, then find out the equation of AB vertical bisector, and then combine 2x-y-3 = 0 to find the center of the circle. But the answer is wrong. Is this method wrong?

The standard equation of a circle passing through points a (5,2) and B (3, - 2) with the center of the circle on the line 2x-y-3 = 0 I will find out the straight line L through AB, then find out the equation of AB vertical bisector, and then combine 2x-y-3 = 0 to find the center of the circle. But the answer is wrong. Is this method wrong?

The midpoint of points a (5,2) and B (3, - 2) is (4,0), and the slope is 2, so
The equation of AB vertical bisector is y = - X / 2 + 2
The intersection point with the line 2x-y-3 = 0 is C (2,1)
|CA|=|CB|=√10=R
The standard equation of circle is (X-2) &# 178; + (Y-1) &# 178; = 10