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I just thought of a much easier way!
We had defined u = z - x, w = y - x
and we have
|z - x|^2 = |z - y|^2
iff
|u|^2 = |u - w|^2
iff
<u,u> = <u,u> - 2<u,w> + <w,w>
iff
2<u,w> = <w,w>
iff
<2u - w, w> = 0
We seek all vectors u which satisfy the above equation, and we also have the constraints
|w| = d
|u| = r
This is really easy to see geometrically. Draw a vector v1 from 0 to w, and another vector v2 from 0 to 2u. Then 2u - w is the vector from the end of v1 to the end of v2.
The condition <2u - w, w> = 0 says precisely that 2u - w is orthogonal to w.
Under what circumstances is this possible?
There are two possibilities: case 1 (sort of a degenerate case) is when 2u = w [in which case 2r = d]
The other possibility is when you can draw a right triangle with a base of length |w| = d and a hypotenuse of length 2|u| = 2r. This is of course possible if and only if 2r > d, and it's equally clear that in that case there are infinitely many choices for u.
We had defined u = z - x, w = y - x
and we have
|z - x|^2 = |z - y|^2
iff
|u|^2 = |u - w|^2
iff
<u,u> = <u,u> - 2<u,w> + <w,w>
iff
2<u,w> = <w,w>
iff
<2u - w, w> = 0
We seek all vectors u which satisfy the above equation, and we also have the constraints
|w| = d
|u| = r
This is really easy to see geometrically. Draw a vector v1 from 0 to w, and another vector v2 from 0 to 2u. Then 2u - w is the vector from the end of v1 to the end of v2.
The condition <2u - w, w> = 0 says precisely that 2u - w is orthogonal to w.
Under what circumstances is this possible?
There are two possibilities: case 1 (sort of a degenerate case) is when 2u = w [in which case 2r = d]
The other possibility is when you can draw a right triangle with a base of length |w| = d and a hypotenuse of length 2|u| = 2r. This is of course possible if and only if 2r > d, and it's equally clear that in that case there are infinitely many choices for u.
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