Consider a rectangular cartesian system and let $b, d, r\in\mathbb R$ such that $d\neq 0$ and $r>0$. The parameter $d$ is allowed to vary, however $b$ and are fixed once they are given.
I'm trying to find the cartesian equation of the curve which consists of all points $(x, y)$ in the plane which are the intersection between the circle centered at $(d, 0)$ with radius $r$ and the straight line through $(0, -b)$ and $(d, 0)$.
Well, the points $(x, y)$ should be such that:
$$(x, y)=(1-t)\cdot (0, -b)+t\cdot (d, 0)$$
for some $t\in \mathbb R$ and
$$(x-d)^2+y^2=r^2.$$
From the first condition, it would follow
$$x=t\cdot d\ \textrm{and}\ y=(t-1)\cdot b.$$
Hence, $$y=\left(\frac{x}{d} -1\right) b=\frac{b}{d}(x-d).$$ Then,
$$r^2=(x-d)^2+\frac{b^2}{d^2}(x-d)^2$$ which implies
$$r^2 d^2=(b^2+d^2)(x-d)^2.$$
There is also an expression involving $y$. In fact, using
$$(x-d)=\frac{d}{b} y$$
it follows
$$\frac{d^2}{b^2} y^2+y^2=r^2$$
and then
$$b^2r^2=(b^2+d^2)y^2$$.
However, I'd like to get a single expression involving $x$ and $y$, since I want to see the curve as the zero set of a polynomial in the variables $x$ and $y$.
What am I missing?
Thanks.
The equation of the line is $$\frac{x}{d}-\frac{y}{b}=1$$ and the equation of the circle is $$(x-d)^2+y^2 = r^2$$
From the first equation $$d=\frac{b\,x}{b+y}$$
Substitute in the second equation to get
$$\left(x-\frac{b\,x}{b+y}\right)^2+y^2=r^2$$
which simplifies to $$y^2(x^2+(b+y)^2) = r^2(b+y)^2$$