CAT 2010 | Question: 12

Question

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

• $x^{2}-y^{2}=0$
• $(x-k)^{2}+y^{2}=1$

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

Answer 1

Given that:
$x^{2}-y^{2}=0\implies y^{2} = x^{2}\rightarrow(1)$

$(x-k)^{2}+y^{2}=1$

$\implies x^{2} + k^{2} – 2kx + x^{2} – 1 = 0$

$\implies 2x^{2}  + (-2k)x  + (k^{2}  – 1) = 0$

For unique solution $:\Delta = 0$

$\implies b^{2} – 4ac = 0$

$\implies (-2k)^{2} – 4(2)(k^{2}  – 1) = 0$

$\implies 4k^{2} – 8(k^{2}  – 1) = 0$

$\implies k^{2} – 2(k^{2}  – 1) = 0$

$\implies k^{2} – 2k^{2}  + 2 = 0$

$\implies k^{2} = 2$

$\implies k = \pm \sqrt{2}$

For positive value of $’x’$, we should take $ k = \sqrt{2}.$

So, the correct answer is $(C).$