Alternatively, you could do it algebraically. Equate the two expressions for y:
x2+k = 1-kx
x2 + kx + k-1 = 0. (1)
The discriminant of this quadratic is k2-4*1*(k-1) or k2-4k+4. This is zero when k = 2.
Since there is only a single solution to equation (1) when the discriminant is zero, k = 2 doesn’t allow the original two equations to intersect in more than one point.