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

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

x² - y² = 0
(x - k)² + y² = 1

Add the equations:
x² + (x - k)² = 1

expand
x² + x² - 2kx + k² = 1
2x² - 2kx + k² - 1 = 0

use the quadratic formula
x = (2k ± √(4k² - 8k² + 8))/4
x = (k ± √(-k² + 2))/2

Since x > 0,This must be greater than 0
(k + √(2 - k²))/2 > 0
k + √(2 - k²) > 0
√(2 - k²) > -k
You can now substitute each choice, and find out that 0 and √2 works.

^_^

Answer 2

x²-y²=0 ==> x²=y² ==> x = y and x = -y so the constraints on the solution are {x,-x}

Since (x-k)²+y² = 1 describes a circle with radius 1 centered at x = k on the x axis, it's pretty obvious (by inspection) that -√2 and 0 are out. If k = √2, then there is no unique solution so it must be that k = 2.


Doug

Answer 3

properly in case you sparkling up for X in the first equation you get X=Y replace X for Y in the 2d equation and think ofyou've got: (X - ok)^2 + X^2 = a million then plug on your accessible solutions to discover such X fee.

Answer 4

x^2-2kx+k^2+y^2=1
x^2-y^2=0 adding
2x^2-2kx+k^2=0
for unique positive solution b^2-4ac=0
4k^2-8k=0
4k^2=8k
4k=8
and so k=2 so choice (a)