Finding a simpler differential equation for this geometric problem This is the problem I am trying to solve:

Find the curves which tangent segment between the coordinate axis is constant

Let $k$ be the length of the segment, then we have to find the points of intersection of the tangent line to the curve with the $x$ and $y$ axis, $Q$ and $P$ respectively. Using the equation $Y-y=y’(Q-x)$, we obtain:
$$
Q = x - \frac{y}{y’}
$$
and
$$
P = y’(\frac{y}{y’}-x)
$$
So the equation for the distance would be
$$
k^2=(y’)^2(\frac{y}{y’}-x)^2+(x - \frac{y}{y’})^2 
$$
I changed it up a little bit and ended up with this
$$
k^2=(y’)^2(\frac{y}{y’}-x)^2+\frac{1}{(y’)^2}(xy’ - y)^2 
$$
I tried expanding the squares, but the equation is too hard for me to solve. I think the reasoning that got me there is right, so I thought there must be a simpler way to express the differential equation. The final equation kinda looks like the one of a circumference, so maybe there’s a way to parametrize it with that, but I couldn’t manage to do anything. I would really appreciate any hint to keep going, since I’ve been stuck with this problem for a couple of days now.

Rearranging we find that we have a Clairaut's equation
\begin{align}
xy'-y=\frac{\pm ky'}{\sqrt{1+(y')^2}},
\end{align}
taking a derivative yields that
\begin{align}
y''\left(x\pm\frac{k}{(1+(y')^2)^{3/2}}\right)=0.
\end{align}
For $y''=0$ we arrive at the general solution
\begin{align}\tag{*}\label{family}
(y-Cx)^2=\frac{(kC)^2}{(1+C^2)}.
\end{align}
For $(1+(y')^2)^{3/2}x\pm k=0$ we find the singular solution is an astroid
\begin{align}\tag{**}\label{envelope}
y^{2/3}+x^{2/3}=k^{2/3},
\end{align}
which is the envelope for the family of lines in equation \ref{family}.
