Rearrangement of polynomial under square root In this page, the following equation
$$\sqrt{(x + c)^2 + y^2} + \sqrt{(x - c)^2 + y^2} = 2a \label{a} \tag{1}$$
is suggested to be arranged in this way:
$$\sqrt{(x + c)^2 + y^2} = 2a - \sqrt{(x - c)^2 + y^2}$$
and then it is squared in both sides. What if instead the square of $\ref{a}$ is directly taken? $x, y$ are real variables and $a,c$ are real constants.
$$(x + c)^2 + y^2 + 2\sqrt{(x + c)^2 + y^2}\sqrt{(x - c)^2 + y^2} + (x - c)^2 + y^2 = 4a^2 \label{b} \tag{2}$$
Focusing on the remaining square root only:
$$\sqrt{(x + c)^2 + y^2}\sqrt{(x - c)^2 + y^2} = \\
= \sqrt{(x^2 + c^2 + y^2 + 2cx)(x^2 + c^2 + y^2 - 2cx)} = \\
= \sqrt{(x^2 + c^2 + y^2)^2 - 4c^2x^2} = \\
= \sqrt{x^4 + c^4 + y^4 + 2c^2x^2 + 2 c^2y^2 + 2x^2y^2 - 4c^2x^2} = \\
= \sqrt{x^4 + c^4 + y^4 - 2c^2x^2 + 2 c^2y^2 + 2x^2y^2}$$
It could still seem the square of $(\pm x^2 \pm c^2 \pm y^2)$, but there is no combination of signs of $x^2,c^2,y^2$ which generates only one negative $- 2c^2x^2$ term and two positive ones as above. How to proceed, then, to simplify it?
Equation $\ref{b}$ should become:
$$(x + c)^2 + y^2 + 4a\sqrt{(x - c)^2 + y^2} - (x - c)^2 - y^2 = 4a^2$$

I think the following way is better.
Since $a>0$ we obtain that our equation is equal to
$$\left(\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}\right)^2=4a^2$$ or
$$\sqrt{(x^2+y^2+z^2)^2-4x^2c^2}=2a^2-x^2-y^2-c^2$$ and since $a>c$ and by the triangle inequality
$$\sqrt{x^2+y^2}<\frac{d_1+d_2}{2}=a,$$ we obtain:
$$2a^2-x^2-y^2-c^2>2a^2-a^2-c^2=a^2-c^2>0,$$ which says that our equation is equivalent to
$$(x^2+y^2+z^2)^2-4x^2c^2=(2a^2-x^2-y^2-c^2)^2$$ and the rest is smooth.
