Metamath Proof Explorer


Theorem itsclc0

Description: The intersection points of a line L and a circle around the origin. (Contributed by AV, 25-Feb-2023)

Ref Expression
Hypotheses itsclc0.i I=12
itsclc0.e E=msup
itsclc0.p P=I
itsclc0.s S=SphereE
itsclc0.0 0˙=I×0
itsclc0.q Q=A2+B2
itsclc0.d D=R2QC2
itsclc0.l L=pP|Ap1+Bp2=C
Assertion itsclc0 ABCA0B0R+0DX0˙SRXLX1=AC+BDQX2=BCADQX1=ACBDQX2=BC+ADQ

Proof

Step Hyp Ref Expression
1 itsclc0.i I=12
2 itsclc0.e E=msup
3 itsclc0.p P=I
4 itsclc0.s S=SphereE
5 itsclc0.0 0˙=I×0
6 itsclc0.q Q=A2+B2
7 itsclc0.d D=R2QC2
8 itsclc0.l L=pP|Ap1+Bp2=C
9 rprege0 R+R0R
10 elrege0 R0+∞R0R
11 9 10 sylibr R+R0+∞
12 11 adantr R+0DR0+∞
13 12 3ad2ant3 ABCA0B0R+0DR0+∞
14 eqid pP|p12+p22=R2=pP|p12+p22=R2
15 1 2 3 4 5 14 2sphere0 R0+∞0˙SR=pP|p12+p22=R2
16 15 eleq2d R0+∞X0˙SRXpP|p12+p22=R2
17 13 16 syl ABCA0B0R+0DX0˙SRXpP|p12+p22=R2
18 fveq1 p=Xp1=X1
19 18 oveq2d p=XAp1=AX1
20 fveq1 p=Xp2=X2
21 20 oveq2d p=XBp2=BX2
22 19 21 oveq12d p=XAp1+Bp2=AX1+BX2
23 22 eqeq1d p=XAp1+Bp2=CAX1+BX2=C
24 23 8 elrab2 XLXPAX1+BX2=C
25 24 a1i ABCA0B0R+0DXLXPAX1+BX2=C
26 17 25 anbi12d ABCA0B0R+0DX0˙SRXLXpP|p12+p22=R2XPAX1+BX2=C
27 18 oveq1d p=Xp12=X12
28 20 oveq1d p=Xp22=X22
29 27 28 oveq12d p=Xp12+p22=X12+X22
30 29 eqeq1d p=Xp12+p22=R2X12+X22=R2
31 30 elrab XpP|p12+p22=R2XPX12+X22=R2
32 3simpa ABCA0B0R+0DABCA0B0
33 32 adantr ABCA0B0R+0DXPABCA0B0
34 simpl3 ABCA0B0R+0DXPR+0D
35 1 3 rrx2pxel XPX1
36 1 3 rrx2pyel XPX2
37 35 36 jca XPX1X2
38 37 adantl ABCA0B0R+0DXPX1X2
39 6 7 itsclc0xyqsol ABCA0B0R+0DX1X2X12+X22=R2AX1+BX2=CX1=AC+BDQX2=BCADQX1=ACBDQX2=BC+ADQ
40 33 34 38 39 syl3anc ABCA0B0R+0DXPX12+X22=R2AX1+BX2=CX1=AC+BDQX2=BCADQX1=ACBDQX2=BC+ADQ
41 40 expcomd ABCA0B0R+0DXPAX1+BX2=CX12+X22=R2X1=AC+BDQX2=BCADQX1=ACBDQX2=BC+ADQ
42 41 expimpd ABCA0B0R+0DXPAX1+BX2=CX12+X22=R2X1=AC+BDQX2=BCADQX1=ACBDQX2=BC+ADQ
43 42 com23 ABCA0B0R+0DX12+X22=R2XPAX1+BX2=CX1=AC+BDQX2=BCADQX1=ACBDQX2=BC+ADQ
44 43 adantld ABCA0B0R+0DXPX12+X22=R2XPAX1+BX2=CX1=AC+BDQX2=BCADQX1=ACBDQX2=BC+ADQ
45 31 44 biimtrid ABCA0B0R+0DXpP|p12+p22=R2XPAX1+BX2=CX1=AC+BDQX2=BCADQX1=ACBDQX2=BC+ADQ
46 45 impd ABCA0B0R+0DXpP|p12+p22=R2XPAX1+BX2=CX1=AC+BDQX2=BCADQX1=ACBDQX2=BC+ADQ
47 26 46 sylbid ABCA0B0R+0DX0˙SRXLX1=AC+BDQX2=BCADQX1=ACBDQX2=BC+ADQ