A circle with radius $5$ and center $(a,b)$ is tangent to the lines $y = 6$ and $y = x.$ Compute the largest possible value of $a + b.$
Let $(x,y)$ be a point on the circle that is tangent to $y = x,$ and let $(x,6)$ be the point of tangency with the line $y = 6,$ as shown below.
[asy]
unitsize(1.5 cm);
real upperparab (real x) {
return ((x + 3)^2/8);
}
pair[] T;
pair tangent = (-1/3,1/3);
T[1] = tangent;
T[2] = tangent + (-3/2,-15/2);
T[3] = tangent + (3/2,-15/2);
draw(reflect((0,0),(1,1))*(parabola(1,0)),red);
draw(reflect((0,0),(1,1))*(parabola(1,0)),red);
draw((0,0)--(0,6)--(T[2])--cycle,blue+linewidth(1.5));
draw(T[2]--T[3]--(0,0),red);
draw((-2,0)--(2,0));
draw(graph(upperparab,-3.5,3.5));
dot("$(0,6)$", (0,6), N);
dot("$(x,y)$", T[2], SW);
label("$y = 6$", (2.2,6), E);
label("$y = x$", (2.8,upperparab(2.8)), E);
[/asy]
Note that $\angle OT_2 L$ is a right angle. The line $y = 6$ has a slope of 0, so $\angle OT_3 E$ is a right angle. Therefore, quadrilateral $T_2 T_3 EL$ is a square, so $OT_2 = OL = T_2 L.$
The circle is tangent to $y = x,$ so $T_2 E$ is perpendicular to $T_2 L.$
[asy]
unitsize(1.5 cm);
real upperparab (real x) {
return ((x + 3)^2/8);
}
pair[] T;
pair tangent = (-1/3,1/3);
T[1] = tangent;
T[2] = tangent + (-3/2,-15/2);
T[3] = tangent + (3/2,-15/2);
draw(reflect((0,0),(1,1))*(parabola(1,0)),red);
draw(reflect((0,0),(1,1))*(parabola(1,0)),red);
draw((0,0)--(0,6)--(T[2])--cycle,blue+linewidth(1.5));
draw(T[2]--T[3]--(0,0),red);
draw((0,0)--(T[2].x,0),dashed);
draw((0,6)--(T[2].x,6),dashed);
draw(T[2]--(T[2].x,0));
draw(T[2]--foot(T[2],(0,0),(T[2].x,6)));
draw(rightanglemark((T[2].x,0),T[2],foot(T[2],(0,0),(T[2].x,6))));
draw(rightanglemark((0,6),T[2],(T[2].x,6)));
dot("$(0,6)$", (0,6), N);
dot("$(x,y)$", T[2], SW);
dot("$L$", foot(T[2],(0,0),(T[2].x,6)), NE);
dot("$(x,0)$", (T[2].x,0), S);
label("$y = 6$", (2.2,6), E);
label("$y = x$", (2.8,upperparab(2.8)), E);
[/asy]
Thus, $EL = T_2 L - ET_2 = OL - OT_2 = 6 - OL.$
Let $(a,b)$ be the center of the circle. Then $(x,6)$ is the point of tangency of this circle with the line $y = 6,$ so the line between $(x,0)$ and $(x,6)$ is perpendicular to the tangent line at $(x,6).$
[asy]
unitsize(1.5 cm);
real upperparab (real x) {
return ((x + 3)^2/8);
}
pair[] T;
pair tangent = (-1/3,1/3), center = (0,-4);
T[1] = tangent;
T[2] = tangent + (-3/2,-15/2);
T[3] = tangent + (3/2,-15/2);
pair circ = centre(T[1],T[2],T[3]);
pair vertex = (circ.x + circ.y)/4;
pair focus = (circ.x - circ.y)/2;
real parab (real x) {
return (c*(x - circ.x)^2 + circ.y);
}
draw(reflect((0,0),(-1,1))*(parabola(1,0)),red);
draw(reflect((0,0),(-1,1))*(parabola(1,0)),red);
draw((0,-upperparab(0))--(0,6)--T[2]--cycle,blue+linewidth(1.5));
draw(center--T[2]--vertex);
draw(T[2]--T[3]--focus);
draw((0,-4.9)--(centre(T[2],T[3],(T[2].x,0)--(T[2].x,6)),dashed);
dot("$(0,6)$", centre(T[2],T[3],(T[2].x,0)--(T[2].x,6)), E);
dot("$(x,y)$", T[2], SW);
dot("$(a,b)$", centre(T[2],T[3],focus));
dot("$(x,0)$", (T[2].x,0), S);
label("$y = 6$", (2.2,6), E);
label("$y = x$", (2.8,upperparab(2.8)), E);
[/asy]
Note that the parabola $y = \frac{(x - a)^2}{5} + b$ is symmetric about the line $y = b.$ Therefore, the vertex is $(a,b),$ so the point of tangency of this parabola and the line $y = 6$ is $(x,6).$ Thus, $(x,6)$ is also the point of tangency of the circle and the parabola, so the tangent line to this parabola at $(x,6)$ passes through $(a,b).$
The equation of the parabola is $y = \frac{(x - a)^2}{5} + b,$ so
\[y = \frac{(x - a)^2}{5} + b + 6.\]Setting this equal to $x,$ we get
\[\frac{(x - a)^2}{5} + b + 6 = x,\]which simplifies to $x^2 - 2ax + 5b - 5x = 0.$ Since the line $y = x$ is tangent to this parabola,
\[\triangle = (-2a)^2 - 4(5b - 5a) = 0,\]so $4a^2 - 20a + 20b = 0.$ This equation can be rewritten as $a^2 - 5a + 5b = 0.$ Then by Vieta's formulas, $a$ is one of the roots of $t^2 - 5t + 5b = 0.$
By the quadratic formula,
\[t = \frac{5 \pm \sqrt{5 - 20b}}{2}.\]Since $t$ is the $x$-coordinate of the center, both roots must be positive. Hence,
\[5 - 20b \ge 0,\]so $b \le \frac{1}{4}.$
The center lies within the triangle whose vertices are $(0,0),$ $(0,5),$ and $(5,0),$ so $a \in [0,5].$ Then $2a - 5 \in [-5,5].$ By Vieta's formulas,
$a + 2a - 5 = 5,$ so $a = \frac{5}{3}.$ Then $b = \frac{5}{3} \cdot \frac{5}{3} = \frac{25}{9},$ so $a + b = \boxed{\frac{70}{9}}.$
Let $P$ be the point $(a,b),$ and let $O$ be the center of the circle.
[asy]
unitsize(0.8 cm);
real upperparab (real x) {
return(x);
}
pair A, B, O, P;
real a, b;
a = 2;
b = 3;
O = (a,b);
P = (a + 5,b);
A = (5,6);
B = (0,0);
draw(graph(upperparab,-1.5,3.5));
draw((-1.5,-1.5)--(3.5,3.5),red);
draw(A--P--B--cycle);
draw(O--P);
dot("$P$", P, SE);
dot("$O$", O, SE);
dot("$A$", A, N);
dot("$B$", B, SW);
[/asy]
Since $O$ lies on the line $y = x,$ $a = b.$ Also, $P$ lies on the line $y = 6,$ so $b = 6.$ Hence, the sum of the coordinates $a + b$ is $a + a = 2a.$
Notice that
\[OP = \sqrt{(a + 5 - a)^2 + (6 - 6)^2} = 5.\]Also, $OP$ makes a right angle with the line $y = 6,$ since the segment from the center of a circle to a tangent line is perpendicular to the tangent line. Hence, $OA = \sqrt{OP^2 - AP^2} = \sqrt{5^2 - 6^2} = \sqrt{-11},$ and
\[a + b = 2a = 2 \sqrt{-11}.\]Since $-11 < 0,$ $\sqrt{-11}$ is purely imaginary. Hence, $2 \sqrt{-11}$ is purely imaginary, so the largest possible value of $a + b$ is $\boxed{0}.$