Let x + y = k.    (1)

Then the quadratic equation represents a circle, while equation (1) represents a straight line.


So, the problem is to find the tangent line (1) to the circle with greatest k.


Or, in algebra language, we want to find k such a way, that quadratic equation and equation (1)
have only one solution (representing the tangent point), which provides highest value of "k".


So, we express from (1)  y = k-x  and substitute it inte the quadratic equation.  We get then

    x^2 + (k-x)^2 - 8x + 6*(k-x) + 23 = 0,

    x^2 + k^2 - 2kx + x^2 - 8x + 6k - 6x + 23 = 0,

    2x^2 - (2k+14)x +(k^2 + 6k + 23) = 0.    (2)


The condition that the circle and the line have only one common point is equivalent 
to the condition that the discriminant of equation (2) is zero.


The discriminant is

    d = b^2 - 4ac = (2k+14)^2 - 4*2*(k^2 + 6k + 23) = 4k^2 + 56k + 106 - 8k^2 - 48k - 184 = 

      = -4k^2 + 8k + 12.


Hence, the condition that the discriminant equal to zero is this quadratic equation for "k"

    4k^2 - 8k - 12 = 0.


Simplify and then factor

    k^2 - 2k - 3 = 0,

    (k-3)*(k+1) = 0.


The roots are k = 3  and k = -1.


We want the greatest "k", so the ANSWER to the problem's question is k = 3.