straight line intersecting a circle
- Thread starter darkyadoo
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MeredithSwift
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Maybe you can use the algebra way to do this, you write out a function for the circle and then you write out a function for a line. You let two function equal and then calculate the number of the points that when the line pass through an interior point of a circle it intersects with that circle. I mean maybe the calculation is a little bit tiring but you can kip lots of thinging methods. All you need to do is calculate!
lev888
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How is interior point defined?
MeredithSwift
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Maybe he means that an arbitrary point in the circle.
lev888
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What does "in the circle" mean? My point, no pun intended, is that to define an interior point you need the notion of distance. One definition I found defines interior points as points whose distance to the center is less than the circle radius.
Sorry for the delay, I was at work...
I work with the axiom of Hilbert (axioms of congruence). Let [imath]\mathcal{P}[/imath] be a plane, for a given segment [imath]\overline{OA}[/imath], we define a cercle [imath]\mathcal{C}[/imath] such that :[math]\mathcal{C}=\big\{B\in \mathcal{P}:\overline{OB}\cong\overline{OA}\big\}[/math]
For the interior of the circle we have [math]\mathring{\mathcal{C}}=\big\{\overline{OX}\subset \mathcal{P} :\overline{OX}\cong\overline{OA}\big\}-\mathcal{C}[/math].
Now I will explain why I ask this question. I wanted to prove the radius-tangent theorem, but something stopped me, I check on youtube to know how they manage the problem, unfortunately they don't.
video
at 5:45 on the video, he supposes that the radius is not perpendicular to the tangent, then he says there exists a segment perpendicular to the tangent which intersects it at X, then he admits that the point B — corresponding to the intersection of the cercle and [imath]\overline{OX}[/imath] — is between O and X . This is my problem ! Because we must prove it. What I propose is to suppose X between O and B. We deduce that X is an interior point of the cercle. Then if we know that :
"a straight line passing through an interior point of a circle intersect it twice"
we have a contradiction, because the tangent of the cercle intersects the cercle only once, so B is between O and X.
I work with the axiom of Hilbert (axioms of congruence). Let [imath]\mathcal{P}[/imath] be a plane, for a given segment [imath]\overline{OA}[/imath], we define a cercle [imath]\mathcal{C}[/imath] such that :[math]\mathcal{C}=\big\{B\in \mathcal{P}:\overline{OB}\cong\overline{OA}\big\}[/math]
For the interior of the circle we have [math]\mathring{\mathcal{C}}=\big\{\overline{OX}\subset \mathcal{P} :\overline{OX}\cong\overline{OA}\big\}-\mathcal{C}[/math].
Now I will explain why I ask this question. I wanted to prove the radius-tangent theorem, but something stopped me, I check on youtube to know how they manage the problem, unfortunately they don't.
video
at 5:45 on the video, he supposes that the radius is not perpendicular to the tangent, then he says there exists a segment perpendicular to the tangent which intersects it at X, then he admits that the point B — corresponding to the intersection of the cercle and [imath]\overline{OX}[/imath] — is between O and X . This is my problem ! Because we must prove it. What I propose is to suppose X between O and B. We deduce that X is an interior point of the cercle. Then if we know that :
"a straight line passing through an interior point of a circle intersect it twice"
we have a contradiction, because the tangent of the cercle intersects the cercle only once, so B is between O and X.