SOLUTION: The graph of (x-3)^2 + (y-5)^2=16 is reflected over the line y=2. The new graph is the graph of the equation x^2 + Bx + y^2 + Dy + F = 0 for some constants B, D, and F. Find B+D+F.

Question 1024954: The graph of (x-3)^2 + (y-5)^2=16 is reflected over the line y=2. The new graph is the graph of the equation x^2 + Bx + y^2 + Dy + F = 0 for some constants B, D, and F. Find B+D+F.
Answer by fractalier(6550) (Show Source): You can put this solution on YOUR website!
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
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