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# Grahping question

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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.

Jun 26, 2021

#2
+115951
+2

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.

Jun 26, 2021

#2
+115951
+2

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.

Melody Jun 26, 2021