+526 Let $x$ and $y$ be real numbers such that $x^2 + y^2 = 4x + 12y.$ Find the largest possible value of $x + y.$ Give your answer in exact form using radicals, simplified as far as possible.
+129649 Simplify as
x^2 -4x + y^2 - 12y = 0 complete the square on x,y
x^2 -4x + 4 + y^2 - 12y + 36 = 4 + 36
(x -2)^2 + (y -6)^2 = 40
This is a circle centered at (2,6) with a radius of sqrt (40)
max x + y can be found by the intersection of the line y =(x -2) + 6 → y = x + 4 with the circle
The intersection point is
x = 2 + sqrt 40 * cos 45° y = 6 + sqrt 40 * sin 45°
max [ x ] + [ y ] = [ 2 + sqrt (40) / sqrt (2)] + [ 6 + sqrt (40)/sqrt (2)] =
8 + 2sqrt (20) =
8 + 4sqrt (5)
