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 $B+D+F$.
I'm confused... The answer is NOT 2.
the first equation is a circle with center at (3,5) and radius 4
it is reflected over y=2
drawing a picture a got
the new center is at (3,-1) radius still 4
that means the equation is (x-3)^2 + (y+1)^2 = 16
expand that to x^2-6x+9 + Y^2 +2Y +1 = 16
combine like terms to get
x^2-6x + Y^2 +2Y -6 so b= -6 d= 2 f= -6
b+d+f = - 10