Solving equations involving square roots

The left hand side of this equation us equivalent to the distance from the point (x, y) and the point (0, 2) added to the distance from the point (x, y) and the point (0, -2). If these distances add to 6, then we can see that the solution forms an ellipse with centre (0, 0) foci (0, 2), (0, -2). By solving for the points where the distance from both of the foci is 3 we get the vertices at $(\sqrt 5, 0)$, $(-\sqrt5, 0)$, $(0, 3)$, $(0, -3)$. So the equation of the ellipse and hence the solution is given by: $$\frac{x^2}{5}+\frac{y^2}{9} = 1$$

Use

$$\sqrt a+\sqrt b=c$$ then

$$a+2\sqrt{ab}+b=c^2$$

then

$$4ab=(c^2-a-b)^2=c^4-2c^2(a+b)+(a+b)^2$$

and finally

$$c^4-2c^2(a+b)+(a-b)^2=0.$$

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In your case,

$$1296-144(x^2+y^2+4)+64y^2=0$$

or

$$\left(\frac x{\sqrt5}\right)^2+\left(\frac y3\right)^2=1,$$ which is a centered, axis-aligned ellipse.

Denote: $$\begin{cases}x^2 + (y-2)^2=t^2 \\ x^2+(y+2)^2=t^2+8y\end{cases}.$$ Then: $$\sqrt{x^2 + (y-2)^2} + \sqrt{x^2 + (y+2)^2} = 6 \Rightarrow \\ t+\sqrt{t^2+8y}=6 \Rightarrow \\ t^2+8y=36-12t+t^2 \Rightarrow \\ t=\frac{9-2y}{3}.$$ Plug it into the first equation: $$x^2+(y-2)^2=\left(\frac{9-2y}{3}\right)^2 \Rightarrow \\ 9x^2+9y^2-36y+36=81-36y+4y^2 \Rightarrow \\ 9x^2+5y^2=45 \Rightarrow \\ \frac{x^2}{5}+\frac{y^2}{9}=1.$$