Rose has a spherical plum of radius $2$ and a spherical watermelon of radius $8$. She builds a glass sphere around the two fruits to contain them, making the sphere as small as possible. When she has done this, she places a golf ball inside the glass sphere that is tangent to the plum and watermelon. What is the largest possible radius of the golf ball?
The radius of the plum is $2,$ so it is tangent to the inside of the glass sphere. Let $O$ be the center of the glass sphere, and let $P$ be the center of the plum, as shown below.
[asy]
unitsize(2 cm);
pair O, P;
O = (0,0);
P = (0,-8);
draw(Circle(O,14));
draw(Circle(P,2));
draw((-8,0)--(8,0));
dot("$O$", O, N);
dot("$P$", P, S);
label("$r$", O--P, S);
label("$8$", (O + P)/2, E);
label("$2$", P--(-2,0), S);
dot((-2,0));
[/asy]
Then $OP = 8 - 2 = 6.$
Since the glass sphere containing the plum is tangent to the plum, the glass sphere has radius 2 more than the plum, so the glass sphere has radius $2 + 2 = 4.$ In other words, the radius of the watermelon is 4. This means that the glass sphere is tangent to the watermelon at a point $W,$ which lies on the line joining $O$ with the center $W$ of the glass sphere.
[asy]
unitsize(2 cm);
pair O, P, W;
O = (0,0);
P = (0,-8);
W = (0,-10);
draw(Circle(O,14));
draw(Circle(P,2));
draw(Circle(W,6));
draw((-8,0)--(8,0));
dot("$O$", O, N);
dot("$P$", P, S);
dot("$W$", W, S);
draw(O--(0,-16),dashed);
dot((-2,0));
[/asy]
Let the radius of the golf ball be $r.$ Since the golf ball is tangent to both the plum and the watermelon, it is also tangent to the inside of the glass sphere. Then the distance from the center of the golf ball to the center of the glass sphere is $4 - r,$ so the distance from the center of the golf ball to the center of the plum is
\[4 + 6 - r = 10 - r.\]Since the golf ball is tangent to the plum at a point $Q$ that lies on the line joining $P$ and $W,$ $PQ = QW,$ and $PW = 2 + r,$ we have
\[PQ + QW = PW.\]Then $r + 2 + r = 4 + 10 - r,$ so $3r = 12,$ or $r = 4.$
Therefore, the largest possible radius of the golf ball is $\boxed{4}.$
Let's consider the situation geometrically.
To find the largest possible radius of the golf ball, we need to maximize the size of the glass sphere that contains both the plum and the watermelon.
First, let's find a plane that is tangent to both the plum and the watermelon. The plane will pass through the center of both fruits.
The centers of the plum and watermelon are symmetrically positioned with respect to each other. To see this, draw a line that connects the centers of the plum and watermelon, and notice that it passes through the point of tangency of the golf ball.
The distance between the centers of the plum and watermelon is $8 + 2 = 10$ (the sum of their radii).
Next, draw a line that connects the center of the plum and the point of tangency of the golf ball.
The length of this line is the difference between the radius of the glass sphere and the radius of the plum: $r - 2$.
By Pythagoras' theorem, the square of the hypotenuse (the line connecting the centers of the plum and the point of tangency of the golf ball) equals the sum of the squares of the other two sides.
$(r - 2)^2 + 10^2 = r^2$
Expanding and simplifying this equation, we have:
$r^2 - 4r + 4 + 100 = r^2$
$4r = 104$
$r = \frac{104}{4} = 26$
Therefore, the largest possible radius of the golf ball is $\boxed{26}$.