Reflection of a sphere?

Question

I am having trouble understanding this concept. I have a sphere with the equation (x-1)^2 + (y+2)^2 + (z-4)^2 = 16

1. equation of reflectance about the x,y plane
2. equation of reflectance about the point (2,1,-2)

Thoughts.
1. My general idea upon resolving this is that since it is about a plane where z is consistant (x,y,0) then the equation for the reflectance will just be (x,y,z) = (u,v,-w)

In turn this will give us the equation of
(x-1)^2 + (y+2)^2 + (z+4)^2 = 16

Since the x and y should be the same and the radius has to be the same, its just the z that is in the opposite side of the plane.

2. Now about a point, i am even more lost here.

Thoughts, i know that the point (2,1,-2) is the mid point between the 2 sphere equations, it is not as basic as the plane question because we are shifting all 3 variables about a point. I know the radius is still going to be 4, just not sure how to relate this point to get the equation of reflectance.

Thanks,

Answer 1

1. I believe your argument is totally correct

The centre of the sphere ≡(1, -2, 4) and its radius is 4
So reflect the centre in the x-y plane (x and y do not change, z changes sign)
You get (1, -2, -4) and the radius, of course, remains unchanged So the reflected sphere is
(x - 1)² + (y + 2)² + (z + 4)² = 16

2. Again your argument is totally correct.
Centre ≡(1, -2, 4) and its radius is 4
Midpoint ≡(2, 1, -2)

Let centre of reflected sphere be (a, b, c)
Then (2, 1, -2) ≡½(1 + a, -2 + b, 4 + c)
Whence (a, b, c) ≡(3, 4, -8)
Again the radius remains unchanged
So the new sphere is:
(x - 3)² + (y - 4)² + (z + 8)² = 16

Answer 2

Think in terms of vectors. Find the offset vector from the point to any point on the sphere. Then the reflection is -1 times that vector since all three coordinates are reflected. Draw a sketch in 2 dim with a circle at (2,5) and reflect it about the point (3,4) to work out the vectors.