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?

If you had actually checked each candidate point for equality of distance from the coordinate axes and $(4,2)$, you would have seen that only $(2,2)$ and $(10,10)$ satisfy the conditions of the original question.
