Question 1192295: 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 $B+D+F$.
Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
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: -10
Answer by greenestamps(13200)  (Show Source):
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