<s>[INST] Solving equations involving square roots I am a student and I often encounter these type of equations: 
$$\sqrt{x^2 + (y-2)^2} + \sqrt{x^2 + (y+2)^2} = 6$$
I usually solve these by taking one term ($\sqrt{x^2 + (y-2)^2}$ for example) to the right hand side but this seems to take more time. Please suggest me some methods which can help me solve these types of problem quickly.
Thanks
 [/INST] A nice way of doing problems like these is taking advantage of our knowledge of the factorization of difference of squares.
Given that
$$\begin{align}\sqrt{x^2 +(y-2)^2} + \sqrt{x^2+(y+2)^2} &= 6 &[1]\\ \end{align}$$
and that
$$\begin{align}(x^2 +(y-2)^2) - (x^2+(y+2)^2) &= -8y &[2]\\ \end{align}$$
we have (by $\frac{[2]}{[1]}$)
$$\begin{align}\sqrt{x^2 +(y-2)^2} - \sqrt{x^2+(y+2)^2} = \frac{-8y}{6} &= -\frac{4}{3}y &[3]\\ \end{align}$$
Adding $[1]$ and $[3]$ gives us
$$\begin{align}
2\sqrt{x^2 +(y-2)^2}  &= 6-\frac{4}{3}y &[4]\\
4(x^2 +(y-2)^2)  &= \bigg(6-\frac{4}{3}y\bigg)^2\\
4x^2 +4y^2-16y+16  &= \frac{16}{9}y^2-16y+36\\
x^2 +y^2+4  &= \frac{4}{9}y^2+9\\
x^2 +\frac{5}{9}y^2  &= 5\\
\frac{x^2}{5} +\frac{y^2}{9}  &= 1\\
\end{align}$$
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