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?
 A: 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.
A: All solutions must lie in the same quadrant as $(4,2)$. If any solution is in a different quadrant, it's distance from both axes is less than distance from $(4,2)$. Can you see why?

 The vector joining any point in a different quadrant cuts the axes before reaching the point $(4,2)$. The length of perpendicular dropped on the closest axis from the solution point is less than the length of this vector (along hypotenuse).

A: Point lies on the bisector $(t,t)$
$$(t-4)^2+(t-2)^2=t^2$$
$$t^2-12 t+20=0\to t=2;\;t=10$$
Points are $(2,2);\;(10,10)$
The other possibility, $A$ on the other bisector $(t,-t)$, leads to the equation:
$$(-t-2)^2+(t-4)^2 =t^2$$
which has no real solutions.
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A: $$\dfrac{x-4}{\cos t}=\dfrac{y-2}{\sin t}=r\ge0$$
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 $p=-2,r=?$
A: (-4,2) (-4,-2) (4,-2)
The original point is 4 units from the y axis and two units from the x axis. i.e distance from each axis. The question is asking to find every point which is 4 units from the y axis AND 2 units from the x axis. There is one such point in each quadrant, which are the three additional points I listed. If you graph it, it should make sense. Those three points are the only ones which will satisfy the condition of distance of 4 from the y axis and distance of 2 from the x axis. It is basically a question about symmetry around each axis.
These represent the remaining three corners of an 8x4 rectangle, centered on the origin.  One point in each quadrant.  I believe most people are overthinking what the problem actually asks for.
