Find the values of k for which the equations x + y = k and x^2 + y^2 = k have a unique solution.
+128475 x + y = k
y = -x + k
x^2 + ( -x + k)^2 = k
x^2 + x^2 - 2xk + k^2 = k
2x^2 - 2kx + k^2 - k = 0
If this has a unique solution for k then
(-2k)^2 - 4 (2) ( k^2 - k ) = 0
4k^2 - 8 k^2 + 8k = 0
8k - 4k^2 = 0
4k ( 2 - k) = 0
k = 0 (trivial)
k = 2
When k = 2
x + y = 2 ⇒ y = 2 - x
x^2 + y^2 = 2
x^2 + (2 -x)^2 = 2
x^2 + x^2 - 4x + 4 = 2
2x^2 - 4x + 2 = 0
x^2 - 2x + 1 = 0
(x - 1)^2 = 0
x = 1
And
x + y = 2
1 + y = 2
y = 1
So....the solution point is (1,1) when k = 2
