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I have the equation

$ \sqrt {(x-a)^2 + y^2 } + \sqrt {(y-a)^2 + x^2 } = |\sqrt {2}a| $

And condition

$ x^2 + y^2 \leq 18$

I need to know for which $a$ the solution is unique, but I don't wanna use geometrical way (pictures and so on, from geometrical perspective it is obvius that $a = 6$, $0$ or $-6$).

I used the fact that solution must have form $(x,x)$ (it cannot be $(x,y)$ for distinct $x$,$y$) and I've got this way $a \in [-6,6]$.

Now I need some kind of sufficient condition to remove all points except of $-6$, $6$ and $0$.

  • 0
    Am I not allowed to use basic geometric conic sections? – 2017-01-07
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    I was hoping to come to some pure algebraic solution without any geometry. I have already solved it geometrical way. – 2017-01-07

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