Q1:

We have, x² = y² and (x − k)² + y² = 1 Solving the two equations simultaneously, we get, (x – k)² + x² = 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=+-root2

Since k is positive the other solution is ruled out k=root2

Hence, option 3.