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 $F$.

Algebra ->  Quadratic-relations-and-conic-sections -> 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 $F$.      Log On


   

Question 1085549: 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 Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The original circle is centered at (3,5) with a radius of 4.
Reflecting across the y=2 shifts the y coordinate of the center.
The center is 3 units above the line y=2.
The reflected circle will have the center 3 units below the line y=2.
2-3=-1
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%28x-3%29%5E2%2B%28y%2B1%29%5E2=16
x%5E2-6x%2B9%2By%5E2%2B2y%2B1=16
highlight%28x%5E2-6x%2By%5E2%2B2y-6=0%29
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I leave it to you to determine the sum B%2BD%2BF
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