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Explanation:

We have, x2 = y2 and (xk) 2 + y2 = 1

Solving the two equations simultaneously, we get,

(xk) 2 + x2 = 1   ...(1)

⇒ x2 – 2kx + k2 + x2 = 1

 2x2 – 2kx + (k2 – 1) = 0

If this equation has a unique solution for x, then discriminant = 0

∴ 4k2 – 8(k2 – 1) = 0

 8 – 4k2 = 0

 k2 = 2

k = ± √2

Now, k = - √​​​​​​​2 will give us a negative solution for (1) and hence it is rejected. [x should be positive according to the question.]

k = √2

Hence, option (c).

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