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I have so far...

Given a line l drawn from a point A on the circle to another point B on the inside of the circle. We want to show that line l will intersect the circle exactly one more time.

I know I need to use theorem 2.25: Given a circle and a point a inside the circle, any ray AB emanating from A must intersect the circle.

I'm just wondering where I should go from my givens.

Lacey Schleicher
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  • What's the set of axioms based on which you are operating? – xyzzyz Sep 23 '20 at 21:22
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    Assuming typical geometric construction rules and the theorem you listed, you can say that segment $\overline{AB}$ is part of a ray $\overrightarrow{BA}$, which we already know intersects the circle (at A). You should now be able to construct the opposite ray, which must also intersect the circle. And the two rays together must form a line. Does that help? – Eric Snyder Sep 23 '20 at 21:49

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Maybe not what you want, but here is an example of using Cartesian co-ordinates to simplify the solving. Choose a co-ordinate system with $A=(0,0),$ with the circle $c$ centered at $O=(r,0).$

So $c=\{(x,y): (x-r)^2+y^2=r^2\}.$ And $B=(u,v)$ with $(u-r)^2+v^2<r^2.$ The line $l$ is $\{(zu,zv):z\in \Bbb R\}.$

Now we have $$(0,0)=A\ne (zu,zv)\in l\cap s \iff (z\ne 0 \land (zu-r)^2+(zv)^2=r^2)\iff $$ $$\iff (z\ne 0 \land z^2u^2-2zur+z^2v^2=0) \iff$$ $$\iff (z\ne 0 \land zu^2-2ur+zv^2=0)\iff$$ $$\iff 0\ne z=\frac {2ur}{u^2+v^2}\iff$$ $$\iff z=\frac {2ur}{u^2+v^2}. $$ The last two steps above are justified by (i) $u^2+v^2\ne 0,$ otherwise $B=(u,v)=(0,0)=A$, and (ii) $u\ne 0,$ otherwise $r^2>(u-r)^2+v^2=(0-r)^2+v^2=r^2+v^2\ge r^2.$

And we have $\frac {2ur}{u^2+v^2}\ne 1,$ otherwise $(u-r)^2+v^2=r^2$, which would imply $B=(u,v)\in c$.

So there is exactly one $C=(zu,zv)\in l\cap s$ with $C\ne A.$

Geometrically, if $B$ is not on the line $m$ thru $A$ and $O$ then let $D$ be on $m$ with $AD\perp DB.$ Then $C$ is on $l,$ with B between A and C, with $\frac {CA}{BA}=|z|=\frac {2|u|r}{u^2+v^2}=\frac {2r\cdot AD}{AD^2+DB^2}.$

DanielWainfleet
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