If a circle in the first quadrant has radius $r$ and is tangent to both coordinate axes, both coordinates of its center can be written in terms of $r$.

The distance between the two centers must equal the sum of the two radii. After solving, use the condition about the center's $x$-coordinate to select the valid radius.

Algebra is the fastest method because the key circle relationship is the distance between centers. Desmos can verify the resulting radius equation.

Enter $y=(10-x)^2+(4-x)^2$ and $y=(x+2)^2$. Here, $x$ represents a possible radius of circle $D$.

Click the intersection points. Their $x$-coordinates are the possible radii. Keep only the coordinate that is less than $10$, as required by the location of the center.

Complete the square to rewrite the equation of circle $C$ as $(x-10)^2+(y-4)^2=4$. Thus, circle $C$ has center $(10,4)$ and radius $2$.

Let the radius of circle $D$ be $r$. Because its center is in the first quadrant and the circle is tangent to both coordinate axes, the center of circle $D$ is $(r,r)$.

For externally tangent circles, the distance between their centers equals the sum of their radii. Therefore, $(10-r)^2+(4-r)^2=(r+2)^2$. Simplifying gives $r^2-32r+112=0$, which factors as $(r-4)(r-28)=0$.

The possible radii are $4$ and $28$. The $x$-coordinate of the center of circle $D$ equals $r$, and it is given to be less than $10$. Therefore, the radius is $4$.