Equation 1 + Equation 2:
$\displaystyle \displaystyle (x^2-y^2) + [(x-a)^2+y^2] = 0+0$
$\displaystyle \displaystyle x^2 + x^2 - 2ax + a^2 = 0$
$\displaystyle \displaystyle 2x^2 - 2ax + a^2 = 0$.
This is a quadratic. For a unique solution, $\displaystyle \displaystyle \Delta = 0$.
$\displaystyle \displaystyle (-2a)^2 - 4(2)(a^2) = 0$
$\displaystyle \displaystyle 4a^2 - 8a^2 = 0$
$\displaystyle \displaystyle -4a^2 = 0$
$\displaystyle \displaystyle a = 0$.
So this system will have a unique solution when $\displaystyle \displaystyle a = 0$.