edited by
2,717 views

1 Answer

0 votes
0 votes
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 $(C).$

Related questions

2 votes
2 votes
1 answer
1
Arjun asked Mar 1, 2020
1,015 views
If $r, s$ and $t$ are consecutive odd integers with $r < s < t$, which of the following must be true?$rs = t$$r + t = 2t – s$$r + s = t – 2$$r + t = 2s$
0 votes
0 votes
0 answers
2
1 votes
1 votes
1 answer
3
Arjun asked Mar 1, 2020
634 views
$\text{P, Q and R}$ are three consecutive odd numbers in ascending order. If the value of three times $\text{P}$ is three less than two times $\text{R}$, find the value o...
1 votes
1 votes
1 answer
4
0 votes
0 votes
0 answers
5