Squaring $\sqrt{(x-2)^{2}+(y-3)^{2}}=\frac{|5x+6y+5|}{\sqrt{5^{2}+6^{2}}}$ does not lose information?

Question

Scenario 1:

$$y=3x+5\tag{1}$$

$$y^{2}=\left(3x+5\right)^{2}\tag{2}$$

When both sides are squared in $(1)$, we lose some information and get $(2)$: the graphs of $(1)$ and $(2)$ aren't the same.

Scenario 2:

$$\sqrt{\left(x-2\right)^{2}+\left(y-3\right)^{2}}=\frac{\left|5x+6y+5\right|}{\sqrt{5^{2}+6^{2}}}\tag{3}$$

$$\left(x-2\right)^{2}+\left(y-3\right)^{2}=\frac{\left(5x+6y+5\right)^{2}}{5^{2}+6^{2}}\tag{4}$$

The graphs of $(3)$ and $(4)$ are exactly the same! When both sides are squared in $(3)$, we get $(4)$, and no information is lost in the process: the two graphs are identical!

Question

1. Why is information preserved in scenario 2, i.e. the graphs of $(3)$ & $(4)$ are identical, when information is lost in scenario 1, i.e. the graphs of $(1)$ & $(2)$ aren't identical?

Answer 1

The equation "$A^2=B^2$" is equivalent to "$A=B$ or $A=-B$".

In your first example, both $y = 3x+5$ and $y = -(3x+5)$ are perfectly possible, and $y^2 = (3x+5)^2$ combines those.

In your second example, both $\sqrt{\left(x-2\right)^{2}+\left(y-3\right)^{2}}$ and $\frac{\left|5x+6y+5\right|}{\sqrt{5^{2}+6^{2}}}$ are guaranteed to be positive, so one cannot be the negative of the other; only the $A=B$ case is a possibility to begin with.

In other words: by squaring both sides, we are adding in all points which satisfy $$\sqrt{\left(x-2\right)^{2}+\left(y-3\right)^{2}} = -\frac{\left|5x+6y+5\right|}{\sqrt{5^{2}+6^{2}}}$$ but there are no such points.

Answer 2

By definition, all the terms in (3) and (4) are already positive: the $\sqrt{}$ argument is the sum of squares and the absolute value term is, by definition, always non-negative.

So there are no negative values for the squaring to make positive and therefore create an extraneous root.

Answer 3

$ |a|=|b|\iff a^2=b^2\kern.6em\not\kern-.6em\implies a=b\implies a^2=b^2.$