Q10
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
Answer: C
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).