# How to prove that there are exactly two points that are on the circle and the line.

Theorem: Let $A$ be a point on the interior of a circle, and let $l$ be a line on $A$. Then there are exactly two points $B, C ∈ l$ that are on the circle.

How do you prove this theorem? Below is what I have so far, which I know is right, but I don't know how to finish the proof? I know I need to show that there cannot be a third point, but how would I show this?

Proof: We first show that there are two points on both circle and line, and then show that there cannot be a third. Let $O$ be the center of the circle, and $r > 0$ be its radius. Let $P \neq A$ be any other point on $l$. Pick a ruler on $l$ so that the coordinates of $A$, $P$ are $0, p > 0$. Let $D ∈ l$ be the point with coordinate $2r$. Then $OA < r$ since $A$ is on the interior of the circle. By Corollary 5.19.C, $AD < OA + OD$, so $AD − OA < OD$. Since $AD = 2r$ and $OA < r$, we have $r < OD$, which means that $D$ is in the exterior of the circle.

Let $f(x)$ represent the distance between $O$ and the point on the ray $→AB$ with coordinate $x$. Since $f(0) < r$ and $f(2r) > r$, there must be some coordinate $b ∈ (0, 2r)$ so that $f(b) = r$ by Theorem 1.3. Then Theorem 4.6 tells us that the point $B$ with coordinate $b$ is on the ray $→AP$. Similarly, there is some point $C$ on the opposite ray having coordinate $c ∈ (−2r, 0)$. This shows that there are two points $B, C$ on both the circle and line.

Definition of Circle: Let $O$ be a point and $r > 0$ a real number. The circle with center $O$ and radius $r$ is the set of points ${A : OA = r}$. We use the term radius in two ways: to refer to a segment $OA$, where $O$ is the center and point $A$ is on the circle, and alternately, to refer to the distance $OA$ of such a segment.

Definition: The interior of the circle with center $O$ and radius $r$ is the set of points ${A : OA < r}$, and the exterior of the circle is the set of points ${A : OA > r}$.

Corollary 5.19.C (Triangle Inequality). In triangle $∆ABC$, $AB + BC ≥ AC$.

Theorem 1.3 (Intermediate Value Theorem). If $f(x)$ is a continuous function on an interval $[a, b]$, and $L$ is any value between $f(a)$ and $f(b)$, then there is some point $c ∈ [a, b]$ so that $f(c) = L$.

Theorem 4.6. Let $A, B$ be points on a line with a ruler so that $a, b$ are their respective coordinates. Let $C$ be any point collinear with $A, B$, and $c$ its coordinate. If $a < b$, then $→AB = {C : a ≤ c}$.

Let's do a proof by contradiction, using nothing more than Euclidean geometry.

Suppose that three points of a line, $B$, $C$, and $D$, lie on a circle centered at $O$. Without loss of generality, assume that $C$ lies between $B$ and $D$. Let $E$ and $F$ be the midpoints of $BC$ and $CD$, respectively. Note that $\angle BEC=\angle ECF=\angle CFD=\pi$, since all five points lie on a line, with $E$ between $B$ and $C$, $C$ between $E$ and $F$, and $F$ between $C$ and $D$.

Since $OB\cong OC$ (because $B$ and $C$ lie on the circle centered at $O$) and $BE\cong CE$ (because $E$ is the midpoint of $BC$), we have $\triangle OEB\cong\triangle OEC$ by the side-side-side theorem (the two triangles sharing the side $OE$), and likewise $\triangle OFC\cong\triangle OFD$. Therefore $\angle OEB=\angle OEC$ and $\angle OFC=\angle OFD$. But $\angle OEB+\angle OEC=\angle BEC=\pi$, so we must have $\angle OEC=\pi/2$, and likewise $\angle OFC=\pi/2$. That is, $\triangle OEC$ and $\triangle OFC$ are both right triangles, with right angle at $E$ and $F$, respectively.

This means that the other angles in $\triangle OEC$ and $\triangle OFC$ are acute. In particular, $\angle OCE\lt\pi/2$ and $\angle OCF\lt\pi/2$. But that implies $\angle ECF=\angle ECO+\angle OCF\lt\pi/2+\pi/2=\pi$, which is a contradiction. So no line can intersect a circle at three points.

Here's an answer from a different universe :

The circle is quadratic and the line is linear. Substituting the linear into the quadratic, the resulting equation is a quadratic which can have at most two roots.

You probably took this problem out of a book? With only the tools you presented, I'm tempted to say it's impossible to give you a satisfying answer. Generally speaking there are many ways to prove the result you want, but it's difficult to guess what tools/definitions/properties would fit in the framework of the book, and thus which one we are "allowed" to use.

You've already proven there are at least $2$ intersection points $B$ and $C$, if you can prove that any point between $B$ and $C$ is in the interior of the circle, you can then easily prove there are no third intersection. Indeed no matter how you arrange those $3$ collinear points, one will always be inside the segment defined by the other $2$, and cannot be on the cicle.

Now as to how you can prove the above property, it depends on what you can use.

• How do you define a line, collinearity, or even distance? Given the level of detail and that you implicity used that your function $f$ is continuous, I'd assume you have a definition somewhere... If you can use a $2D$ coordinate system, it'd probably be easier to use the equations of a line and circle, and use marty cohen's suggestion instead of mine.
• Are you allowed to use vectors and norms? Then for any $D$ on segment $BC$ there exists $\lambda$, $0<\lambda< 1$ such that $\overrightarrow{OD}=\lambda\overrightarrow{OB}+(1-\lambda)\overrightarrow{OC}$. You have $$OD=\lVert \overrightarrow{OD}\rVert \le \lambda\lVert\overrightarrow{OB}\rvert +(1-\lambda)\lVert \overrightarrow{OC}\rVert =\lambda OB +(1-\lambda)OC < r$$
• Can you use Pythagoras and orthogonal projections? Consider $O'$ the orthogonal projection of $O$ onto the line. For any other point $D$ of the line, the triangle $OO'D$ has a right angle at point $O'$, you can then apply Pythagoras, deduce relations on the various side lengths and those suffice to conclude. (If $O$ belongs to the line you can't apply Pythagoras because the triangle is flat and $0'=0$... But in that case $0$ has a coordinate on your "ruler" and it should be even easier to conclude.)
• Other methods...

Now some nitpicking on your proof. You use the triangular inequality to get "$AD < OA + OD$", it should have been "$AD \le OA + OD$". The inequality is strict if and only if $O$, $A$ and $D$ are not collinear, in other words only if triangle $OAD$ is not flat. Your conclusion still holds because $OA < r$ is indeed a strict inequality.

Inversive geometry viewpoint:

Let our circle be called $C$. Invert around the center of $C$. We know that intersections are preserved with inversions, so we can just inspect the result of this inversion. $C$ will remain unchanged, while with the line, there are two cases. In the case it intersects the center, the 2 solutions are trivial, since $l$ makes a diameter. If not, it will become a circle through the center of our circle. Let this circle be called $C_i$

Since there exists a point through $l$ that is in the interior of the circle (by hypothesis), it must invert to a point outside of the circle, and we conclude that $C_i$ must contain points outside of the circle. That is, our circle does not contain $C_i$

Therefore, they must either intersect twice, or $C$ and $C_i$ must be tangent. If $C_i$ was tangent to $C$, then we could conclude that $l$ is tangent to $C$, but then there would be no points lying on the line within the interior of the circle; a contradiction! Thus, they must intersect twice.