A circle with center $C$ is tangent to the positive $x$ and $y$-axes and externally tangent to the circle centered at $(3,0)$ with radius $1$. What is the sum of all possible radii of the circle with center $C$?
Let $r$ be the radius of such a circle.  Since the circle is tangent to the positive $x$-axis and positive $y$-axis, its center is $(r,r).$  This circle is also tangent to the circle centered at $(3,0)$ with radius 1, so
\[(r - 3)^2 + r^2 = (r + 1)^2.\]This simplifies to $r^2 - 8r + 8 = 0.$  By the quadratic formula, the roots are $r = 4 \pm 2 \sqrt{2}.$  Thus, the sum of all possible values of $r$ is $\boxed{8}.$

[asy]
unitsize(1 cm);

pair[] O;
real[] r;

r[1] = 4 - 2*sqrt(2);
O[1] = (r[1],r[1]);
r[2] = 4 + 2*sqrt(2);
O[2] = (r[2],r[2]);

draw(Circle(O[1],r[1]));
draw(arc(O[2],r[2],160,290));
draw(Circle((3,0),1));
draw((-0.5,0)--(9,0));
draw((0,-0.5)--(0,9));
draw(O[1]--(r[1],0));
draw(O[1]--(0,r[1]));
draw(O[1]--(3,0));
draw(O[2]--(r[2],0));
draw(O[2]--(0,r[2]));
draw(O[2]--(3,0));

dot("$(3,0)$", (3,0), S);
dot("$O_1$", O[1], N);
dot("$O_2$", O[2], NE);
[/asy]
8
Level 4
Intermediate Algebra
Problem: A circle with center $C$ is tangent to the positive $x$ and $y$-axes and externally tangent to the circle centered at $(3,0)$ with radius $1$. What is the sum of all possible radii of the circle with center $C$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the radius of the circle with center $C$, we denote the radius as $r$. Since the circle is tangent to the positive $x$ and $y$-axes, the center $C$ must be located at the point $(r, r)$. 

Next, we consider the external tangency condition with the circle centered at $(3, 0)$ with radius $1$. The distance between the centers of the two circles must equal the sum of their radii. The distance between the center $C$ at $(r, r)$ and the center of the other circle at $(3, 0)$ is given by the distance formula:

Distance = $\sqrt{(r - 3)^2 + (r - 0)^2} = \sqrt{(r - 3)^2 + r^2}$.

This distance must equal the sum of the radii, which is $r + 1$. Therefore, we set up the equation:

$\sqrt{(r - 3)^2 + r^2} = r + 1$.

Squaring both sides to eliminate the square root gives:

$(r - 3)^2 + r^2 = (r + 1)^2$.

Expanding both sides results in:

$(r^2 - 6r + 9) + r^2 = (r^2 + 2r + 1)$.

Combining like terms leads to:

$2r^2 - 6r + 9 = r^2 + 2r + 1$.

Rearranging this equation yields:

$r^2 - 8r + 8 = 0$.

Next, we apply the quadratic formula to solve for $r$:

$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} = \frac{8 \pm \sqrt{64 - 32}}{2} = \frac{8 \pm \sqrt{32}}{2} = \frac{8 \pm 4\sqrt{2}}{2} = 4 \pm 2\sqrt{2}$.

This gives us two possible radii: $4 + 2\sqrt{2}$ and $4 - 2\sqrt{2}$. 

To find the sum of all possible radii, we calculate:

$(4 + 2\sqrt{2}) + (4 - 2\sqrt{2}) = 4 + 2\sqrt{2} + 4 - 2\sqrt{2} = 8$.

Thus, the final answer is $8$.<|END_OF_LATENT|> The final answer is $8$.