There is probably a much easier way to do this, but i solved using algebra on the conditions that need to be fulfilled.
We know that there must be a distance of 4 between the point k,2 and x,y. We can relate these values by going (k-x)^2 + (2-y)^2 = 4^2
We know that y = x+4
thus this simplifies to (k-x)^2 + (-2-x)^2 = 16
We also know that the closest distance the lines and the point will be will be at a bisector to the line we are given, this has the negative inverse gradient to the line which = -1.
y - y1 = m (x - x1)
y1 is 2 and x1 is k also y = x + 4
x + 4 - 2 = -1(x - k)
x + 2 + =-x + k
thus k = 2x + 2
substitute this into (k-x)^2 + (-2-x)^2 = 16
we get (x+2)^2 + (x+2)^2 = 16 => 2(x^2) + 8x - 8 = 0
quadratic solvable for x. gives x = 0.8284, -4.828
substitute these into k = 2x + 2
giving 3.6568 and - 7.656