<s>[INST] Find every point whose distance from each of the two coordinate axes equals its distance from the point $(4, 2)$ Here is my attempt:

Let $(x, y)$ be any such point.


Then the distance of $(x, y)$ from the $x$-axis is $\lvert y \rvert$, whereas the distance of that point from the $y$-axis is $\lvert x \rvert$. Thus we have the equalities
$$
\lvert x \rvert = \lvert y \rvert = \sqrt{ (x-4)^2+(y-2)^2}.
$$
From $\lvert x \rvert = \lvert y \rvert$, we obtain $y = \pm x$.

Thus we have the equations
$$
\sqrt{ (x-4)^2 + (\pm x -2)^2} = \lvert x \rvert, 
$$
which implies
$$
(x-4)^2 + ( \pm x - 2)^2 = x^2, 
$$
which simplifies to
$$
x^2 - 2(4 \pm 2)x + 20 = 0.
$$

Thus we have the following two quadratic equations
$$
x^2 - 12x + 20 = 0 \qquad \mbox{ and } \qquad x^2 -4x + 20 = 0,
$$
and the solutions of these quadratic equations are
$$
x = \frac{12 \pm 8 }{ 2 } \qquad \mbox{ and } \qquad x = \frac{ 4 \pm 8 \iota }{ 2 },
$$
that is,
$$
x = 10, 2 \qquad \mbox{ and } \qquad x = 2 \pm 4 \iota.
$$
We will of course only need the real values for our $x$.


Thus there are eight possible points satisfying the condition given in the problem, namely
$$
(10, 10), (10, -10), (-10, 10), (-10, -10), (2, 2), (2, -2), (-2, 2), (-2, -2). 
$$

Is my solution correct in terms of the technique employed as well as the answers obtained? Or, are there any mistakes?
 [/INST] $$\dfrac{x-4}{\cos t}=\dfrac{y-2}{\sin t}=r\ge0$$
So, the point $P(4+r\cos t,2+r\sin t)$$
We need $$(4+r\cos t)^2=(2+r\sin t)^2=r^2$$
$$r(1+\cos t)=4$$ as $r(1-\cos t)\ge0$
Similarly, $r(\sin t+1)=2$
Using https://en.m.wikipedia.org/wiki/Weierstrass_substitution#The_substitution, $$\dfrac42=\dfrac2{(1+p)^2}$$
where $p=\tan\dfrac t2$
$\implies(1+p)^2=1\implies p=0,-2$
If $p=0,r=2$
If $p=-2,r=?$
</s>