Question

For which value of $k$ does the pair of equations yield a unique positive $x$ solution? \[ x^2 - y^2 = 0 \] \[ (x - k)^2 + y^2 = 1 \]

• A. 2
• B. 0
• C. $\sqrt{2}$
• D. $-\sqrt{2}$

Hint:

Combine equations systematically and apply discriminant conditions for uniqueness.

Answer 1

The Correct Option is C

From $x^2 - y^2 = 0$, we have $y = \pm x$. Case $y = x$: $(x - k)^2 + x^2 = 1$.
Case $y = -x$: $(x - k)^2 + x^2 = 1$ (same equation).
Simplifies: $2x^2 - 2kx + k^2 - 1 = 0$. For unique $x$: discriminant = 0:
$(-2k)^2 - 4(2)(k^2 - 1) = 0 \Rightarrow 4k^2 - 8k^2 + 8 = 0 \Rightarrow k^2 = 2 \Rightarrow k = \pm \sqrt{2}$.
Positive $x$ condition selects $k = \sqrt{2}$. \[ \boxed{\sqrt{2}} \]