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For which value of $k$ does the following pair of equations yield a unique solution for $x$ such that the solution is positive?

• $x^{2}-y^{2}=0$
• $(x-k)^{2}+y^{2}=1$
1. $2$
2. $0$
3. $\sqrt{2}$
4. $\sqrt{-2}$

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 $(C).$
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