Metamath Proof Explorer


Theorem itsclc0b

Description: The intersection points of a (nondegenerate) line through two points and a circle around the origin. (Contributed by AV, 2-May-2023) (Revised by AV, 14-May-2023)

Ref Expression
Hypotheses itsclc0.i ⊢ I = 1 2
itsclc0.e ⊢ E = I
itsclc0.p ⊢ P = ℝ I
itsclc0.s ⊢ S = Sphere ⁡ E
itsclc0.0 ⊢ 0 ˙ = I × 0
itsclc0.q ⊢ Q = A 2 + B 2
itsclc0.d ⊢ D = R 2 ⁢ Q − C 2
itsclc0.l ⊢ L = p ∈ P | A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C
Assertion itsclc0b ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ A ≠ 0 ∨ B ≠ 0 ∧ R ∈ ℝ + ∧ 0 ≤ D → X ∈ 0 ˙ S R ∧ X ∈ L ↔ X ∈ P ∧ X ⁡ 1 = A ⁢ C + B ⁢ D Q ∧ X ⁡ 2 = B ⁢ C − A ⁢ D Q ∨ X ⁡ 1 = A ⁢ C − B ⁢ D Q ∧ X ⁡ 2 = B ⁢ C + A ⁢ D Q

Proof

Step Hyp Ref Expression
1 itsclc0.i ⊢ I = 1 2
2 itsclc0.e ⊢ E = I
3 itsclc0.p ⊢ P = ℝ I
4 itsclc0.s ⊢ S = Sphere ⁡ E
5 itsclc0.0 ⊢ 0 ˙ = I × 0
6 itsclc0.q ⊢ Q = A 2 + B 2
7 itsclc0.d ⊢ D = R 2 ⁢ Q − C 2
8 itsclc0.l ⊢ L = p ∈ P | A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C
9 rprege0 ⊢ R ∈ ℝ + → R ∈ ℝ ∧ 0 ≤ R
10 elrege0 ⊢ R ∈ 0 +∞ ↔ R ∈ ℝ ∧ 0 ≤ R
11 9 10 sylibr ⊢ R ∈ ℝ + → R ∈ 0 +∞
12 11 adantr ⊢ R ∈ ℝ + ∧ 0 ≤ D → R ∈ 0 +∞
13 12 3ad2ant3 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ A ≠ 0 ∨ B ≠ 0 ∧ R ∈ ℝ + ∧ 0 ≤ D → R ∈ 0 +∞
14 eqid ⊢ p ∈ P | p ⁡ 1 2 + p ⁡ 2 2 = R 2 = p ∈ P | p ⁡ 1 2 + p ⁡ 2 2 = R 2
15 1 2 3 4 5 14 2sphere0 ⊢ R ∈ 0 +∞ → 0 ˙ S R = p ∈ P | p ⁡ 1 2 + p ⁡ 2 2 = R 2
16 15 eleq2d ⊢ R ∈ 0 +∞ → X ∈ 0 ˙ S R ↔ X ∈ p ∈ P | p ⁡ 1 2 + p ⁡ 2 2 = R 2
17 13 16 syl ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ A ≠ 0 ∨ B ≠ 0 ∧ R ∈ ℝ + ∧ 0 ≤ D → X ∈ 0 ˙ S R ↔ X ∈ p ∈ P | p ⁡ 1 2 + p ⁡ 2 2 = R 2
18 fveq1 ⊢ p = X → p ⁡ 1 = X ⁡ 1
19 18 oveq2d ⊢ p = X → A ⁢ p ⁡ 1 = A ⁢ X ⁡ 1
20 fveq1 ⊢ p = X → p ⁡ 2 = X ⁡ 2
21 20 oveq2d ⊢ p = X → B ⁢ p ⁡ 2 = B ⁢ X ⁡ 2
22 19 21 oveq12d ⊢ p = X → A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = A ⁢ X ⁡ 1 + B ⁢ X ⁡ 2
23 22 eqeq1d ⊢ p = X → A ⁢ p ⁡ 1 + B ⁢ p ⁡ 2 = C ↔ A ⁢ X ⁡ 1 + B ⁢ X ⁡ 2 = C
24 23 8 elrab2 ⊢ X ∈ L ↔ X ∈ P ∧ A ⁢ X ⁡ 1 + B ⁢ X ⁡ 2 = C
25 24 a1i ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ A ≠ 0 ∨ B ≠ 0 ∧ R ∈ ℝ + ∧ 0 ≤ D → X ∈ L ↔ X ∈ P ∧ A ⁢ X ⁡ 1 + B ⁢ X ⁡ 2 = C
26 17 25 anbi12d ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ A ≠ 0 ∨ B ≠ 0 ∧ R ∈ ℝ + ∧ 0 ≤ D → X ∈ 0 ˙ S R ∧ X ∈ L ↔ X ∈ p ∈ P | p ⁡ 1 2 + p ⁡ 2 2 = R 2 ∧ X ∈ P ∧ A ⁢ X ⁡ 1 + B ⁢ X ⁡ 2 = C
27 18 oveq1d ⊢ p = X → p ⁡ 1 2 = X ⁡ 1 2
28 20 oveq1d ⊢ p = X → p ⁡ 2 2 = X ⁡ 2 2
29 27 28 oveq12d ⊢ p = X → p ⁡ 1 2 + p ⁡ 2 2 = X ⁡ 1 2 + X ⁡ 2 2
30 29 eqeq1d ⊢ p = X → p ⁡ 1 2 + p ⁡ 2 2 = R 2 ↔ X ⁡ 1 2 + X ⁡ 2 2 = R 2
31 30 elrab ⊢ X ∈ p ∈ P | p ⁡ 1 2 + p ⁡ 2 2 = R 2 ↔ X ∈ P ∧ X ⁡ 1 2 + X ⁡ 2 2 = R 2
32 31 anbi1i ⊢ X ∈ p ∈ P | p ⁡ 1 2 + p ⁡ 2 2 = R 2 ∧ X ∈ P ∧ A ⁢ X ⁡ 1 + B ⁢ X ⁡ 2 = C ↔ X ∈ P ∧ X ⁡ 1 2 + X ⁡ 2 2 = R 2 ∧ X ∈ P ∧ A ⁢ X ⁡ 1 + B ⁢ X ⁡ 2 = C
33 anandi ⊢ X ∈ P ∧ X ⁡ 1 2 + X ⁡ 2 2 = R 2 ∧ A ⁢ X ⁡ 1 + B ⁢ X ⁡ 2 = C ↔ X ∈ P ∧ X ⁡ 1 2 + X ⁡ 2 2 = R 2 ∧ X ∈ P ∧ A ⁢ X ⁡ 1 + B ⁢ X ⁡ 2 = C
34 simpl1 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ A ≠ 0 ∨ B ≠ 0 ∧ R ∈ ℝ + ∧ 0 ≤ D ∧ X ∈ P → A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ
35 simpl2 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ A ≠ 0 ∨ B ≠ 0 ∧ R ∈ ℝ + ∧ 0 ≤ D ∧ X ∈ P → A ≠ 0 ∨ B ≠ 0
36 simpl3l ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ A ≠ 0 ∨ B ≠ 0 ∧ R ∈ ℝ + ∧ 0 ≤ D ∧ X ∈ P → R ∈ ℝ +
37 simpl3r ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ A ≠ 0 ∨ B ≠ 0 ∧ R ∈ ℝ + ∧ 0 ≤ D ∧ X ∈ P → 0 ≤ D
38 1 3 rrx2pxel ⊢ X ∈ P → X ⁡ 1 ∈ ℝ
39 1 3 rrx2pyel ⊢ X ∈ P → X ⁡ 2 ∈ ℝ
40 38 39 jca ⊢ X ∈ P → X ⁡ 1 ∈ ℝ ∧ X ⁡ 2 ∈ ℝ
41 40 adantl ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ A ≠ 0 ∨ B ≠ 0 ∧ R ∈ ℝ + ∧ 0 ≤ D ∧ X ∈ P → X ⁡ 1 ∈ ℝ ∧ X ⁡ 2 ∈ ℝ
42 36 37 41 jca31 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ A ≠ 0 ∨ B ≠ 0 ∧ R ∈ ℝ + ∧ 0 ≤ D ∧ X ∈ P → R ∈ ℝ + ∧ 0 ≤ D ∧ X ⁡ 1 ∈ ℝ ∧ X ⁡ 2 ∈ ℝ
43 6 7 itsclc0xyqsolb ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ A ≠ 0 ∨ B ≠ 0 ∧ R ∈ ℝ + ∧ 0 ≤ D ∧ X ⁡ 1 ∈ ℝ ∧ X ⁡ 2 ∈ ℝ → X ⁡ 1 2 + X ⁡ 2 2 = R 2 ∧ A ⁢ X ⁡ 1 + B ⁢ X ⁡ 2 = C ↔ X ⁡ 1 = A ⁢ C + B ⁢ D Q ∧ X ⁡ 2 = B ⁢ C − A ⁢ D Q ∨ X ⁡ 1 = A ⁢ C − B ⁢ D Q ∧ X ⁡ 2 = B ⁢ C + A ⁢ D Q
44 34 35 42 43 syl21anc ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ A ≠ 0 ∨ B ≠ 0 ∧ R ∈ ℝ + ∧ 0 ≤ D ∧ X ∈ P → X ⁡ 1 2 + X ⁡ 2 2 = R 2 ∧ A ⁢ X ⁡ 1 + B ⁢ X ⁡ 2 = C ↔ X ⁡ 1 = A ⁢ C + B ⁢ D Q ∧ X ⁡ 2 = B ⁢ C − A ⁢ D Q ∨ X ⁡ 1 = A ⁢ C − B ⁢ D Q ∧ X ⁡ 2 = B ⁢ C + A ⁢ D Q
45 44 pm5.32da ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ A ≠ 0 ∨ B ≠ 0 ∧ R ∈ ℝ + ∧ 0 ≤ D → X ∈ P ∧ X ⁡ 1 2 + X ⁡ 2 2 = R 2 ∧ A ⁢ X ⁡ 1 + B ⁢ X ⁡ 2 = C ↔ X ∈ P ∧ X ⁡ 1 = A ⁢ C + B ⁢ D Q ∧ X ⁡ 2 = B ⁢ C − A ⁢ D Q ∨ X ⁡ 1 = A ⁢ C − B ⁢ D Q ∧ X ⁡ 2 = B ⁢ C + A ⁢ D Q
46 33 45 bitr3id ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ A ≠ 0 ∨ B ≠ 0 ∧ R ∈ ℝ + ∧ 0 ≤ D → X ∈ P ∧ X ⁡ 1 2 + X ⁡ 2 2 = R 2 ∧ X ∈ P ∧ A ⁢ X ⁡ 1 + B ⁢ X ⁡ 2 = C ↔ X ∈ P ∧ X ⁡ 1 = A ⁢ C + B ⁢ D Q ∧ X ⁡ 2 = B ⁢ C − A ⁢ D Q ∨ X ⁡ 1 = A ⁢ C − B ⁢ D Q ∧ X ⁡ 2 = B ⁢ C + A ⁢ D Q
47 32 46 bitrid ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ A ≠ 0 ∨ B ≠ 0 ∧ R ∈ ℝ + ∧ 0 ≤ D → X ∈ p ∈ P | p ⁡ 1 2 + p ⁡ 2 2 = R 2 ∧ X ∈ P ∧ A ⁢ X ⁡ 1 + B ⁢ X ⁡ 2 = C ↔ X ∈ P ∧ X ⁡ 1 = A ⁢ C + B ⁢ D Q ∧ X ⁡ 2 = B ⁢ C − A ⁢ D Q ∨ X ⁡ 1 = A ⁢ C − B ⁢ D Q ∧ X ⁡ 2 = B ⁢ C + A ⁢ D Q
48 26 47 bitrd ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ C ∈ ℝ ∧ A ≠ 0 ∨ B ≠ 0 ∧ R ∈ ℝ + ∧ 0 ≤ D → X ∈ 0 ˙ S R ∧ X ∈ L ↔ X ∈ P ∧ X ⁡ 1 = A ⁢ C + B ⁢ D Q ∧ X ⁡ 2 = B ⁢ C − A ⁢ D Q ∨ X ⁡ 1 = A ⁢ C − B ⁢ D Q ∧ X ⁡ 2 = B ⁢ C + A ⁢ D Q