$$\sqrt{\frac{\left(x-2\right)^{2}+\left(y-3\right)^{2}}{\left(x+1\right)^{2}+\left(y-4\right)^{2}}}=\frac{2}{3}…(i)$$
$$5x^{2}+5y^{2}-44x-22y+49=0…(ii)$$
(i) is the original equation, and (ii) is the simplified version of (i), which I have found by squaring both sides of (i). Even though (ii) is the squared version of (i), the graphs of the two equations are identical. Why is this the case? Why doesn't (ii) have extraneous values?
This might help you in answering the question.
Best Answer
The reason is simple. The equation states that the square root of some quantity equals $2/3$. If you square both sides, it just means that the quantity must equal $4/9$. Nothing has changed, because the only way the LHS can equal $2/3$ is if we take the positive root of the quantity inside the root.
Another way to put it is to consider a simple example: $$\sqrt{z} = 3.$$ In this case, what is the solution set for $z$? It cannot be anything else besides $9$. There is no introduction of an extraneous solution.
Then, why is it that extraneous solution are introduced for, say, $$z = 9 \implies z^2 = 81?$$ In this case, squaring both sides introduces the solution $z = -9$ because the act of squaring lets us choose a negative value, which when squared, becomes positive. But this is not permitted in the square root situation above, since $\sqrt{-9} \ne 3$.