Question 1024951
The graph of
(x-3)^2 + (y-5)^2 = 16
is a circle centered at (3,5) with radius of 4.
When you reflect over the line y = 2, it is still a circle of radius 4 but its new center is at (3,-1), and its new equation is
(x-3)^2 + (y+1)^2 = 16
Now multiply everything out and get
x^2 - 6x + 9 + y^2 + 2y + 1 = 16
x^2 - 6x + y^2 + 2y - 6 = 0
B+D+F = -10