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Question:

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

x² – y² = 0

(x – k) ² + y² = 1

Explanation:

We have, x² = y² and (x − k) ² + y² = 1

Solving the two equations simultaneously, we get,

(x – k) 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).