$x^{2}-y^{2}=0\dots(1)$
$(x-k)^{2}+y^{2}=1\dots(2)$
From (1) and (2)
$(x-k)^{2}+x^{2}=1$
$x^{2}+k^{2}-2kx+x^{2}=1$
$2x^{2}-2kx+k^{2}-1=0$
We knows that for Unique solution of the equation $b^{2}-4ac=0$
here, $4k^{2}-8(k^{2}-1)=0$
$8-4k^{2}=0$
$k^{2}=2$
k=$\pm $$\sqrt 2$
Since k is positive the other solution is ruled out.
Hence,k=$\sqrt 2$
Hence,Option(3)$\sqrt 2$.