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Given that:

$x^{2}-y^{2}=0\implies y^{2} = x^{2}\rightarrow(1)$

$(x-k)^{2}+y^{2}=1$

$\implies x^{2} + k^{2} – 2kx + x^{2} – 1 = 0$

$\implies 2x^{2} + (-2k)x + (k^{2} – 1) = 0$

For unique solution $:\Delta = 0$

$\implies b^{2} – 4ac = 0$

$\implies (-2k)^{2} – 4(2)(k^{2} – 1) = 0$

$\implies 4k^{2} – 8(k^{2} – 1) = 0$

$\implies k^{2} – 2(k^{2} – 1) = 0$

$\implies k^{2} – 2k^{2} + 2 = 0$

$\implies k^{2} = 2$

$\implies k = \pm \sqrt{2}$

For positive value of $’x’$, we should take $ k = \sqrt{2}.$

So, the correct answer is $(C).$

$x^{2}-y^{2}=0\implies y^{2} = x^{2}\rightarrow(1)$

$(x-k)^{2}+y^{2}=1$

$\implies x^{2} + k^{2} – 2kx + x^{2} – 1 = 0$

$\implies 2x^{2} + (-2k)x + (k^{2} – 1) = 0$

For unique solution $:\Delta = 0$

$\implies b^{2} – 4ac = 0$

$\implies (-2k)^{2} – 4(2)(k^{2} – 1) = 0$

$\implies 4k^{2} – 8(k^{2} – 1) = 0$

$\implies k^{2} – 2(k^{2} – 1) = 0$

$\implies k^{2} – 2k^{2} + 2 = 0$

$\implies k^{2} = 2$

$\implies k = \pm \sqrt{2}$

For positive value of $’x’$, we should take $ k = \sqrt{2}.$

So, the correct answer is $(C).$