Metamath Proof Explorer


Theorem 2sphere

Description: The sphere with center M and radius R in a two dimensional Euclidean space is a circle. (Contributed by AV, 5-Feb-2023)

Ref Expression
Hypotheses 2sphere.i ⊢ I = 1 2
2sphere.e ⊢ E = I
2sphere.p ⊢ P = ℝ I
2sphere.s ⊢ S = Sphere ⁡ E
2sphere.c ⊢ C = p ∈ P | p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2 = R 2
Assertion 2sphere ⊢ M ∈ P ∧ R ∈ 0 +∞ → M S R = C

Proof

Step Hyp Ref Expression
1 2sphere.i ⊢ I = 1 2
2 2sphere.e ⊢ E = I
3 2sphere.p ⊢ P = ℝ I
4 2sphere.s ⊢ S = Sphere ⁡ E
5 2sphere.c ⊢ C = p ∈ P | p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2 = R 2
6 prfi ⊢ 1 2 ∈ Fin
7 1 6 eqeltri ⊢ I ∈ Fin
8 simpl ⊢ M ∈ P ∧ R ∈ 0 +∞ → M ∈ P
9 elrege0 ⊢ R ∈ 0 +∞ ↔ R ∈ ℝ ∧ 0 ≤ R
10 9 simplbi ⊢ R ∈ 0 +∞ → R ∈ ℝ
11 10 adantl ⊢ M ∈ P ∧ R ∈ 0 +∞ → R ∈ ℝ
12 eqid ⊢ dist ⁡ E = dist ⁡ E
13 2 3 12 4 rrxsphere ⊢ I ∈ Fin ∧ M ∈ P ∧ R ∈ ℝ → M S R = p ∈ P | p dist ⁡ E M = R
14 7 8 11 13 mp3an2i ⊢ M ∈ P ∧ R ∈ 0 +∞ → M S R = p ∈ P | p dist ⁡ E M = R
15 9 biimpi ⊢ R ∈ 0 +∞ → R ∈ ℝ ∧ 0 ≤ R
16 15 ad2antlr ⊢ M ∈ P ∧ R ∈ 0 +∞ ∧ p ∈ P → R ∈ ℝ ∧ 0 ≤ R
17 sqrtsq ⊢ R ∈ ℝ ∧ 0 ≤ R → R 2 = R
18 16 17 syl ⊢ M ∈ P ∧ R ∈ 0 +∞ ∧ p ∈ P → R 2 = R
19 18 eqeq2d ⊢ M ∈ P ∧ R ∈ 0 +∞ ∧ p ∈ P → p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2 = R 2 ↔ p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2 = R
20 1 3 rrx2pxel ⊢ p ∈ P → p ⁡ 1 ∈ ℝ
21 20 adantl ⊢ M ∈ P ∧ p ∈ P → p ⁡ 1 ∈ ℝ
22 1 3 rrx2pxel ⊢ M ∈ P → M ⁡ 1 ∈ ℝ
23 22 adantr ⊢ M ∈ P ∧ p ∈ P → M ⁡ 1 ∈ ℝ
24 21 23 resubcld ⊢ M ∈ P ∧ p ∈ P → p ⁡ 1 − M ⁡ 1 ∈ ℝ
25 24 resqcld ⊢ M ∈ P ∧ p ∈ P → p ⁡ 1 − M ⁡ 1 2 ∈ ℝ
26 1 3 rrx2pyel ⊢ p ∈ P → p ⁡ 2 ∈ ℝ
27 26 adantl ⊢ M ∈ P ∧ p ∈ P → p ⁡ 2 ∈ ℝ
28 1 3 rrx2pyel ⊢ M ∈ P → M ⁡ 2 ∈ ℝ
29 28 adantr ⊢ M ∈ P ∧ p ∈ P → M ⁡ 2 ∈ ℝ
30 27 29 resubcld ⊢ M ∈ P ∧ p ∈ P → p ⁡ 2 − M ⁡ 2 ∈ ℝ
31 30 resqcld ⊢ M ∈ P ∧ p ∈ P → p ⁡ 2 − M ⁡ 2 2 ∈ ℝ
32 25 31 readdcld ⊢ M ∈ P ∧ p ∈ P → p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2 ∈ ℝ
33 24 sqge0d ⊢ M ∈ P ∧ p ∈ P → 0 ≤ p ⁡ 1 − M ⁡ 1 2
34 30 sqge0d ⊢ M ∈ P ∧ p ∈ P → 0 ≤ p ⁡ 2 − M ⁡ 2 2
35 25 31 33 34 addge0d ⊢ M ∈ P ∧ p ∈ P → 0 ≤ p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2
36 32 35 jca ⊢ M ∈ P ∧ p ∈ P → p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2 ∈ ℝ ∧ 0 ≤ p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2
37 36 adantlr ⊢ M ∈ P ∧ R ∈ 0 +∞ ∧ p ∈ P → p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2 ∈ ℝ ∧ 0 ≤ p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2
38 resqcl ⊢ R ∈ ℝ → R 2 ∈ ℝ
39 sqge0 ⊢ R ∈ ℝ → 0 ≤ R 2
40 38 39 jca ⊢ R ∈ ℝ → R 2 ∈ ℝ ∧ 0 ≤ R 2
41 10 40 syl ⊢ R ∈ 0 +∞ → R 2 ∈ ℝ ∧ 0 ≤ R 2
42 41 ad2antlr ⊢ M ∈ P ∧ R ∈ 0 +∞ ∧ p ∈ P → R 2 ∈ ℝ ∧ 0 ≤ R 2
43 sqrt11 ⊢ p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2 ∈ ℝ ∧ 0 ≤ p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2 ∧ R 2 ∈ ℝ ∧ 0 ≤ R 2 → p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2 = R 2 ↔ p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2 = R 2
44 37 42 43 syl2anc ⊢ M ∈ P ∧ R ∈ 0 +∞ ∧ p ∈ P → p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2 = R 2 ↔ p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2 = R 2
45 8 anim1ci ⊢ M ∈ P ∧ R ∈ 0 +∞ ∧ p ∈ P → p ∈ P ∧ M ∈ P
46 2nn0 ⊢ 2 ∈ ℕ 0
47 eqid ⊢ 𝔼 hil ⁡ 2 = 𝔼 hil ⁡ 2
48 47 ehlval ⊢ 2 ∈ ℕ 0 → 𝔼 hil ⁡ 2 = 1 … 2
49 46 48 ax-mp ⊢ 𝔼 hil ⁡ 2 = 1 … 2
50 fz12pr ⊢ 1 … 2 = 1 2
51 50 1 eqtr4i ⊢ 1 … 2 = I
52 51 fveq2i ⊢ 1 … 2 = I
53 49 52 eqtri ⊢ 𝔼 hil ⁡ 2 = I
54 2 53 eqtr4i ⊢ E = 𝔼 hil ⁡ 2
55 1 oveq2i ⊢ ℝ I = ℝ 1 2
56 3 55 eqtri ⊢ P = ℝ 1 2
57 54 56 12 ehl2eudisval ⊢ p ∈ P ∧ M ∈ P → p dist ⁡ E M = p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2
58 45 57 syl ⊢ M ∈ P ∧ R ∈ 0 +∞ ∧ p ∈ P → p dist ⁡ E M = p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2
59 58 eqcomd ⊢ M ∈ P ∧ R ∈ 0 +∞ ∧ p ∈ P → p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2 = p dist ⁡ E M
60 59 eqeq1d ⊢ M ∈ P ∧ R ∈ 0 +∞ ∧ p ∈ P → p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2 = R ↔ p dist ⁡ E M = R
61 19 44 60 3bitr3d ⊢ M ∈ P ∧ R ∈ 0 +∞ ∧ p ∈ P → p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2 = R 2 ↔ p dist ⁡ E M = R
62 61 rabbidva ⊢ M ∈ P ∧ R ∈ 0 +∞ → p ∈ P | p ⁡ 1 − M ⁡ 1 2 + p ⁡ 2 − M ⁡ 2 2 = R 2 = p ∈ P | p dist ⁡ E M = R
63 5 62 eqtr2id ⊢ M ∈ P ∧ R ∈ 0 +∞ → p ∈ P | p dist ⁡ E M = R = C
64 14 63 eqtrd ⊢ M ∈ P ∧ R ∈ 0 +∞ → M S R = C