The graph of (x - 3)^2 + (y - 5)^2 = 16 is reflected over the line y = x + 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.
+118667 The graph of (x - 3)^2 + (y - 5)^2 = 16 This is a circle centre (3,5) radius 4
is reflected over the line y = x + 2.
I will find the perpendicular distance of y=x+2 to (3,5)
x-y+2=9 (3,5)
\(d=\frac{|3*1+5*-1+2|}{\sqrt{1^2+1^2}}=0\)
So the line goes through the centre of the circle.
So you are refecting the circle over a diameter. So what does this mean?
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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.
+118667 The graph of (x - 3)^2 + (y - 5)^2 = 16 This is a circle centre (3,5) radius 4
is reflected over the line y = x + 2.
I will find the perpendicular distance of y=x+2 to (3,5)
x-y+2=9 (3,5)
\(d=\frac{|3*1+5*-1+2|}{\sqrt{1^2+1^2}}=0\)
So the line goes through the centre of the circle.
So you are refecting the circle over a diameter. So what does this mean?
--------------------
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.
