Unique positive solution for system of equations

From $x^2 - y^2 = 0$, $y^2 = x^2$. Substitute into circle equation: $(x - k)^2 + x^2 = 1 \implies 2x^2 - 2kx + k^2 - 1 = 0$. For a unique solution of $x$, discriminant $D = 0$: $(-2k)^2 - 4(2)(k^2 - 1) = 0 \implies 4k^2 - 8k^2 + 8 = 0 \implies 4k^2 = 8 \implies k = \pm \sqrt{2}$. When $k = \sqrt{2}$, equation is $2x^2 - 2\sqrt{2}x + 1 = 0 \implies (\sqrt{2}x - 1)^2 = 0 \implies x = \frac{1}{\sqrt{2}} > 0$. When $k = -\sqrt{2}$, $x = -\frac{1}{\sqrt{2}} < 0$, which is rejected. Thus $k = \sqrt{2}$.