Given a circle $C$ of radius $1$ centered at the origin, I want to determine the locus of the centers of all circles that touch $C$ and the $x$-axis. This is the red curve in the following Desmos plot, where the blue circle touches $C$ and the $x$-axis:
Let $P=(\sin\alpha,\cos\alpha)$ be the point where the blue circle touches $C$. Moving $\ell$ units towards the center of $C$ (the origin) gives the point $Q=(1-\ell)(\sin\alpha,\cos\alpha)$. Now a circle around $Q$ touches the $x$-axis when the $y$-coordinate of $Q$ equals $\ell$, so that $Q$ has the same distance to $C$ and to $y=0$:
$$
Q_y=(1-\ell)\cos\alpha \stackrel.= \ell \tag1
$$
This equation is solved by
$$Q_y=\frac{\cos\alpha}{1+\cos\alpha} \tag2$$
It's also easy to compute the $x$-ccordinate of $Q$, which yields $Q$ depending on $\alpha$:
$$
Q=Q(\alpha)=\left(\frac{\sin\alpha}{1+\cos\alpha}, \frac{\cos\alpha}{1+\cos\alpha}\right) \tag3
$$
Where I am stuck is to compute $Q_y$ as a function of $Q_x$, that is find $f$ such that
$$
Q_y = f(Q_x) \tag4
$$
By looking at the plot I guessed
$$
f(t) = \frac12(1-t^2) \tag 5
$$
and indeed $Q$ satisfies $(5)$. It's not surprising that $f$ is a curve or order 2. But how to do it without guessing? I have no idea how to find the inverse of
$$
\alpha \mapsto \frac{\sin\alpha}{1+\cos\alpha}
$$
Trying to substitute $\alpha = \arcsin z$ gives
$$
Q(z) = \left(\frac z{1+\sqrt{1-z^2}} , \frac{\sqrt{1-z^2}}{1+\sqrt{1-z^2}} \right)\tag6
$$
just makes it more complicated…
Best Answer
The radius of a blue circle is $y$. The distance between the centers is $1-y$ (do you see why?). Therefore, $x^2 + y^2 = (1-y)^2$.