Yahoo Answers: Answers and Comments for Prove for each r in R , Ar={(x,y): x^2 + y^2 = r^2} is a partition? [Mathematics]
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From Jason
enUS
Wed, 21 Mar 2012 10:40:11 +0000
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Yahoo Answers: Answers and Comments for Prove for each r in R , Ar={(x,y): x^2 + y^2 = r^2} is a partition? [Mathematics]
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https://answers.yahoo.com/question/index?qid=20120321104011AAnV0YP
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From kb: Since r^2 = (r)^2, we can assume without loss...
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https://answers.yahoo.com/question/index?qid=20120321104011AAnV0YP
Wed, 21 Mar 2012 21:17:20 +0000
Since r^2 = (r)^2, we can assume without loss of generality that r ≥ 0.
For each fixed r ≥ 0, A(r) is a circle centered at (0, 0) with radius r.
(When r = 0, then A(0) is simply {(0, 0)}.)
Hence, {A(r) : r ≥ 0} is a family of concentric circles (same center at (0, 0)).
Equivalence relation:
(x, y) ~ (u, v) <==> x^2 + y^2 = u^2 + v^2.
(Under this relation, {A(r) : r ≥ 0} is the collection of distinct equivalence classes.)

Proof that A(r) is a partition.
(i) Given (a, b) in R^2, note that (a, b) is in A(r), where r = √(a^2 + b^2).
Hence, (a, b) is in A(r) for some r ≥ 0.
(ii) Now, we show that A(r) ∩ A(s) = ∅ for distinct r, s ≥ 0.
If (a, b) is in A(r) ∩ A(s), then (a, b) is in both A(r) and A(s).
==> r = √(a^2 + b^2) = s, which contradicts r and s being distinct.

I hope this helps1