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

  • $x^2 - y^2 =0$
  • $(x-k)^2 + y^2 =1$
  1. $2$
  2. $0$
  3. $\sqrt{2}$
  4. -$\sqrt{2}$
in Quantitative Aptitude edited by
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$x^{2}-y^{2}=0\dots(1)$

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

From (1) and (2)

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

$x^{2}+k^{2}-2kx+x^{2}=1$

$2x^{2}-2kx+k^{2}-1=0$

We knows that for Unique solution of the equation $b^{2}-4ac=0$

here, $4k^{2}-8(k^{2}-1)=0$

$8-4k^{2}=0$

$k^{2}=2$

k=$\pm $$\sqrt 2$

Since k is positive the other solution is ruled out.

Hence,k=$\sqrt 2$

 

Hence,Option(3)$\sqrt 2$.
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