Question 1192295
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The circle template in general is
(x-h)^2 + (y-k)^2 = r^2
with (h,k) as the center and r = radius.


For this particular problem
(x-3)^2+(y-5)^2 = 16
we have (h,k) = (3,5) as the center and r = 4 as the radius.


The point (3,5) is exactly 3 units above the horizontal line y = 2
When reflecting over this line, we'll move down 3 units to arrive at the mirror line itself, then move another 3 units further south to get to (3,-1). 
Overall we moved the center down 6 units.
The radius stays the same because lengths and distances are preserved under reflection operations. 


Therefore we go from 
(x-3)^2+(y-5)^2 = 16
to
(x-3)^2+(y+1)^2 = 16
after reflecting the circle over the horizontal line y = 2.


Let's expand things out and get everything to one side.
(x-3)^2+(y+1)^2 = 16
x^2-6x+9+y^2+2y+1 = 16
x^2-6x+9+y^2+2y+1-16 = 0
x^2-6x+y^2+2y-6 = 0


Comparing that last equation to
x^2 + Bx + y^2 + Dy + F = 0
shows that
B = -6
D = 2
F = -6


So,
B+D+F = -6+2+(-6) = -10



Answer: <font color=red>-10</font>
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