I have a Question (although it is a geometry question, I am just putting the part where I am stuck) where: $$6 = \sqrt{(x+3)^2+y^2} = \sqrt{(x-3)^2+y^2}$$
And I have to find the value of $x$ and $y$. Now I know that I can solve: $$6 = \sqrt{(x+3)^2+y^2}$$ and upon doing just that I find:$$x^2+y^2+6x=27 \tag{1}\label{eq1}$$
Now I can also solve: $$6 = \sqrt{(x-3)^2+y^2}$$ By doing that I find: $$x^2+y^2-6x=27\tag{2}\label{eq2}$$
Now I can subtract equation 1 from equation 2 from which I get : $$x = 0$$ Now I can just put the value of $x$ in either equation 1 or equation 2 to find the value of $y$,which comes out to be $$y = \pm \sqrt{27}$$
But what I am actually stuck on is when I try to solve:$$\sqrt{(x+3)^2+y^2} = \sqrt{(x-3)^2+y^2}$$ $$(x+3)^2+y^2 = (x-3)^2+y^2$$ $$(x+3)^2=(x-3)^2$$ $$x^2+9+6x = x^2+9-6x$$ $$6x = -6x$$ $$1=-1$$ Now I don't know what I did wrong here and that's why I am asking here. As always any help to point my dumb a** in the right direction would be much appreciated.
