Deducing a cissoid equation using an intersecting linear equation (Stillwell, 2001) Exercise 2.5.2 in Mathematics and it's History 2nd ed. by John Stillwell asks to reader to deduce the equation for a cissoid using the linear equation;
$$Y=\frac{\sqrt{1-x^2}}{1+x}(X-1)$$
The line, cissoid, and unit circle are all shown in a diagram. The equation for this line was found in the previous question I determined the equation above using $mx+c$. Given, that the line intersects with The unit circle at $-x$ and at $(1,0)$.
Some things can be determined about the cissoid and the points of intersection. These are shown by the diagram. Namely, it's cusp $R$ is at $(1,0)$ and it intersects with the line above at point $P (x,mx+c)$. It's equation should be;
$$y^2(1+x)=({1-x})^3$$
If it wasn't obvious I'm new to studying maths at this level. I've only just learnt what a cissoid is and I'm not sure how these all relate. I could arrive at the answer using other means but it felt like cheating.
Any help would be appreciated,
Thanks,
Henry
 A: I managed to find a copy of Stillwell, Third Edition, online (not sayin' where) and am posting the diagram you refer to.  (I will remark that this is not the standard way that the construction of the cissoid is described, which makes having this essential.  Trying to adapt the construction generally applied, as can be found in numerous sources, made matters rather confusing.)

I am presuming in this approach that the "vertical" lines at $ \ x \ $ and $ \ -x \ $ are allowed to vary within the unit circle and that we are defining points $ \ P \ $ on the cissoid as the intersections of the oblique line and the vertical line at $ \ x \ \ . $  (The more typical derivation produces a curve that is infinite in extent.)
The oblique line meets the circle at $ \ R \ (1 \ , \ 0 ) \ $ and at $ \ -x \ \ , $ where the $ \ y-$coordinate is $ \ -\sqrt{1-x^2} \ \ $ (since the intersection is on the "lower" semi-circle of $ \ x^2 + y^2 \ = \ 1 \ \ .  $ So the slope of this line is $$ \ m \ = \ \frac{y \ - \ 0}{(-x) \ - \ 1} \ \ = \ \ \frac{-\sqrt{1-x^2} \ - \ 0}{(-x) \ - \ 1} \ \ = \ \ \frac{\sqrt{1-x^2}}{1 \ + \ x } \ \ . $$
From this, it follows that a line with this slope passing through the point $ \ R \ $ has the equation
$$ ( y \ - \ 0 ) \ \ = \ \ \frac{\sqrt{1-x^2}}{1 \ + \ x } · (x \ -  \ 1 ) \ \ . $$
This is the result for Exercise 2.5.1.
Stillwell now labels a point on the cissoid as $ \ P \ (X \ , \ Y) \ . $  So the locus of points where the vertical and oblique lines intersect is given by the equation $ \ Y \ \ = \ \ \frac{\sqrt{1-X^2}}{1 \ + \ X } · (X \ -  \ 1 ) \ \ . $  The cissoid equation is then found by algebraic manipulation:
$$ \ Y \ · \ (1 \ + \ X) \ \ = \ \ \sqrt{1-X^2} \ · \ (X \ -  \ 1 ) \  \ \ \Rightarrow \ \ Y^2 \ · \ (1 \ + \ X)^2 \ \ = \ \ (1 \ - \ X^2) \ · \ (X \ -  \ 1 )^2 $$
$$ \Rightarrow \ \ Y^2 \ · \ (1 \ + \ X)^2 \ \ = \ \ (1 \ + \ X) \ · \ (1 \ - \ X) \ · \ (1 \ -  \ X )^2 $$
$$ \Rightarrow \ \ Y^2 \ · \ (1 \ + \ X) \ \ = \ \ (1 \ -  \ X )^3  \ \ , $$
with $ \ X \ \neq \ -1 \ \ ; $ we thus have the equation sought in Exercise 2.5.2 .
It should be said that when we squared the equation in the derivation, we lifted the restriction on using only the lower semi-circle, so this equation in fact describes a curve symmetrical about the $ \ x-$axis.  A graph of the equation (along with its asymptote; see below) shows that it extends indefinitely beyond the unit circle; this is what is found from the more usual constructions, which you will see elsewhere.
Since Diocles was interested in using this curve for a very specific application (the "duplication of the cube" problem), though, the portion Stillwell describes was likely sufficient.  (Indeed, symbolic algebra and curve-plotting appeared over a millennium after Diocles.)
[Note in passing: while the part of the cissoid within the unit circle suggests an astroid, they are significantly different, with the astroid being rather more concave.]

