# Show that any line through the origin of a circle has two point on the circle and that a circle has infinite points.

Given two distinct points O,A we define the circle with center O and radius OA to be the set $\Gamma$ of all points B such that $OA \cong OB$ .

(a) Show that any line through a meets the circle in exactly two points.

Consider. Pick any point P draw line OP clearly for any choice of P the line OP describes all possible lines passing through the point O.

Consider any line $\ell$ passing through the origin of the the circle $\Gamma$ by I2 any line contains at least 2 distinct points. One is O the other let us denote it P any point on the line different from O.

Then Consider the ray OP it contains a point inside the circle $\Gamma$ Namely O and a we can use Axiom C1 to cut off a line segment off this infite ray of length OA on the ray OP Denote it $p_1$ then $p_1$ lies on the circle $\Gamma$ by definition as $Op_1 \cong OA$

Extend the ray OP to a line then by axiom B2 there exists D s.t $p_1*O*D$ consider the ray OD again by Axiom C1 we cut off a point of length OA on the ray OD denote it $p_2$ then $p_2$ lies on the circle $\Gamma$ by definition as $Op_2 \cong OB$ and clearly $p_1 \neq p_2$ as $D*O*p_1$ and $p_2 * D * O$ by construction hence by exercise we have $p_2 * O * p_1$

Consider the line OP then D and O are both on this line hence by I1 Any two points determine a unique line $\ell$ hence we have that $p_1$ and $p_2$ are on $\ell$ where $\ell$ is any given line

I believe this is correct in the sense that there are at least two points on a circle for any given line passing through the origin but is missing the uniqueness of the two points. Can anyone fix my proof or given an entirely different one that shows the desired result?

Perhaps we can get it by contradiction by assuming there is three points?

(b) Show that a circle contains infinitely many points.

By using above this problem is simplified all we need to show it that given an arbitrary point O that there are an infinite number of unique lines passing through O. if so then there are $2(\infty )$ points on the circle.

EDIT:Perhaps i have this.

BWOC: Consider there are a finite number of unique lines passing through O, since there is a finite number draw them all. now there are infinite number of lines hence there must be a line not passing through O call it $\ell$ as $\ell$ has an infinite number of points. so it can only intersect a finite number of these lines as well there is only finitely many of them. Hence there exists some point P on $\ell$ s.t P is not the point of intersection with any of the finite number of lines Hence P is not on any of the lines. Draw line OP but OP is unique by I1. Hence there is an infinite number of unique lines passing through O each of which by above has at least two points on the circle $\Gamma$ hence $\Gamma$ has an infinite number of points.

Any help much appreciated.

This is Ex 8.5 in Geometry by Hartshorne.

The following axioms and anything provable from them is all i have at my disposal.

Incidence axioms

I1)For every two points A and B there exists a line a that contains them both.

I2)There exist at least two points on a line.

I3)There exist at least three points that do not lie on the same line.

Axioms of betweeness

B1) If a point B lies between points A and C, B is also between C and A, and there exists a line containing the distinct points A, B, C.

B2) If A and C are two points, then there exists at least one point B on the line AC such that C lies between A and B.

B3) Of any three points situated on a line, there is no more than one which lies between the other two.

B4) Pasch's Axiom: Let A, B, C be three points not lying in the same line and let a be a line lying in the plane ABC and not passing through any of the points A, B, C. Then, if the line a passes through a point of the segment AB, it will also pass through either a point of the segment BC or a point of the segment AC.

Axioms of congruence

$C_1$. Given a line segment AB, and given a ray r originating at a point C, there exists a unique point D on the ray r such that $AB \cong CD$.

$C_2$. If $AB \cong CD$ and $AB \cong EF$, then $CD \cong EF$. Every line segment is congruent to itself.

$C_3$. (Addition). Given three points $A$, $B$, $C$ on a line satisfying $A * B * C$, and three further points D, E, F on a line satisfying $D * E * F$, if $AB \cong DE$ and $BC \cong EF$, then $AC \cong DF$.

• If you going to downvote please say why so that i might fix your concern. Thanks. – Faust Nov 26 '17 at 19:09
• I do not personally agree w the downvote but it might be because the formatting makes your question hard to answer. What exactly are you asking? Are you asking for a proof check, for a more formal proof, or for direction on how to complete your proof? – PhilTheLawyer Nov 26 '17 at 19:36
• Well the proof i have given isn't a proof as i have shown that there are at least 2 points. Any proof showing exactly two points whether its an entirely different method or an extension of mine is fine as long as it follows the rules. – Faust Nov 26 '17 at 19:55
• I'm weary of the term "ray". Is it an axiom or definition that given a point A on a line that the line consists of exactly two rays originating at A in opposite directions? I think you need something of the sort. So there is by C1 a point on the ray that creates a segment congruent to a radius, and thus the point is on the circle. By a combination of BETWEEN and CONGRUENT axioms you can prove there are no other such points on the ray. Do the same for the other ray. – fleablood Nov 26 '17 at 20:10
• Exactly! My concern (not being familiar with your text) was how legitimate is it to say "Given a point A, construct the ray ..." But if you can do it, the result is directly by B1. Also in glancing at B1) I didn't realize B1) included uniqueness. I figured you could prove unique via betweeness and a segment not being allowed to be congruent to a smaller segment of itself but I guess it goes the other way around. You use uniqueness to prove a segment can not be congruent to a smaller segment of itself. – fleablood Nov 26 '17 at 20:32