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.
Some helpful info…
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 $→AB = {C : a ≤ c}$.
Best Answer
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.