Prove: Poincare Model I-1 Holes for Every Two Points of E

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Incidence Postualte I-1 holes for the Poincare Model: Every two points of E lie on exactly one L-Line.
Prove: Given any two points P and Q inside the unit circle C, there exists a unique L-line l containing them. (this will require the use of analytic geometry.)

poincare.jpg


L-lines:arcs of circles perpendicular to the unit circle in S and the diameter of S.

How would i solve this? I know that i need to prove and P,Q are not equal to the orgin and that either one is at the orgin, but how?

Thanks for the help!
 
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mathstudent88 said:
Incidence Postualte I-1 holes for the Poincare Model: Every two points of E lie on exactly one L-Line.
Prove: Given any two points P and Q inside the unit circle C, there exists a unique L-line l containing them. (this will require the use of analytic geometry.)

poincare.jpg


L-lines:arcs of circles perpendicular to the unit circle in S and the diameter of S.

How would i solve this? I know that i need to prove and P,Q are not equal to the orgin and that either one is at the orgin, but how?

Thanks for the help!

You don't prove that neither P nor Q is at the origin- there is no reason for that to be true. However, a "hyperbolic line" in the Poincare circle model is either
1) A Euclidean line through the center of the bounding circle (here the origin) or
2) A circle orthogonal to the bounding circle.

IF is P or Q is the origin, then it is easy: the Euclidean line from P to Q is a hyperbolic line. You should also be able to prove, with Analytic geometry, that a circle passing through the origin cannot be orthogonal to the unit circle.

If neither P nor Q are at the origin, then you need to show that there is exactly one circle through both P and Q that is orthogonal to the unit circle. How you would do that, I can't say because I don't know what concepts you know that you can use. I myself would use the fact that the inverse points to P and Q must also be on that circle. Do you know what an "inverse point" in this sense is?
 
HallsofIvy said:
You don't prove that neither P nor Q is at the origin- there is no reason for that to be true. However, a "hyperbolic line" in the Poincare circle model is either
1) A Euclidean line through the center of the bounding circle (here the origin) or
2) A circle orthogonal to the bounding circle.

IF is P or Q is the origin, then it is easy: the Euclidean line from P to Q is a hyperbolic line. You should also be able to prove, with Analytic geometry, that a circle passing through the origin cannot be orthogonal to the unit circle.

If neither P nor Q are at the origin, then you need to show that there is exactly one circle through both P and Q that is orthogonal to the unit circle. How you would do that, I can't say because I don't know what concepts you know that you can use. I myself would use the fact that the inverse points to P and Q must also be on that circle. Do you know what an "inverse point" in this sense is?

Do you mean like P' and Q' are inverse points?
 
HallsofIvy said:
IF is P or Q is the origin, then it is easy: the Euclidean line from P to Q is a hyperbolic line. You should also be able to prove, with Analytic geometry, that a circle passing through the origin cannot be orthogonal to the unit circle.
QUOTE]


To prove this, i would just have to show x^2+y^2+ax+by+1=0, given the points P and Q?
 
mathstudent88 said:
Do you mean like P' and Q' are inverse points?
I have no idea because I have no idea what YOU mean by P' and Q'.
 
mathstudent88 said:
HallsofIvy said:
IF is P or Q is the origin, then it is easy: the Euclidean line from P to Q is a hyperbolic line. You should also be able to prove, with Analytic geometry, that a circle passing through the origin cannot be orthogonal to the unit circle.
QUOTE]


To prove this, i would just have to show x^2+y^2+ax+by+1=0, given the points P and Q?
Again, since you have not said what a and b are, I have no idea what you are talking about!
 
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