Reflection Equation for Sphere in (x-1)^2+(y+2)^2+(z-4)^2=16 at (2,1,-2)

[bobsmiters]

Find the equation of reflection of the sphere in (x-1)^2+(y+2)^2+(z-4)^2=16 with respect to the point (2,1,-2).

There was another question asking for the reflection equation but it was with respect to the xy-plane so it just meant changing some signs. What I came up with for this question was

(x-3)^2+(y-4)^2+(z+8)^2=16 ?

 

[HallsofIvy]

Yes, that is correct.

"Reflection in a point", q, means that the image of point p is on the line from p through q and the same distance from q as p. That is, the "reflection point" must be the the midpoint of the segment between a point and its image. The center of the sphere is at (1, -2, 4). If we take (a, b, c) as the image, then we must have (1+ a)/2= 2, (-2+ b)/2= 1, and (c+ 4)/2= -2. Solving those equations gives a=3, b= 4, c= -8 so your equation is correct. (Of course, the image circle has the same radius as the original.)

 

[Gza]

what exactly does this have to do with calculus?