3
$\begingroup$

As per the definition given on this site,

Given an origin O and a point P on the curve, let B be the point where the extension of the line OP intersects the line $x=2a$ and C be the intersection of the circle of radius $a$ and center $(a,0)$ with the extension of OP. Then the Cissoid of Diocles is the curve which satisfies OP=CB.

enter image description here

But is this condition applicable to the points on the curve that lie outside the circle? Clearly, if point 'P' was outside the circle, then the extension of OP wouldn't have intersected the circle.

What is it that I'm missing here?

$\endgroup$
1

1 Answer 1

2
$\begingroup$

$OP$ is considered part of its own extension, so for $P$ outside the circle, $C$ is between $O$ and $P$ (and the required equality may be restated as $OC=PB$). There is thus no problem in the definition.

$\endgroup$
3
  • $\begingroup$ I get your first point — that the point $C$ will be between $O$ and $P$, but are you sure that the required equality has to be restated to $OC =PB$? Wouldn't the original condition $OP = CB$ still work? $\endgroup$
    – Sasikuttan
    Sep 28, 2022 at 12:06
  • 1
    $\begingroup$ @Curiouserandcuriouser Yes, I never said the original won't work, only that it may be restated that way (which does not involve overlapping line segments). $\endgroup$ Sep 28, 2022 at 12:10
  • $\begingroup$ Okay, I get it now. $OP = CB \iff OC = PB$. $\endgroup$
    – Sasikuttan
    Sep 28, 2022 at 12:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .