Circles with centers $(-3,7)$ and $(2,17)$ have radii $5$ and $10$, respectively. The equation of a common external tangent to the circles can be written in the form $y=mx+b$ with $m>0$. What is $b$?

1. Both centers lie above the tangent line $y=mx+b$, at distances equal to the radii. With $k=\sqrt{1+m^2}$, the distance from $(x_0,y_0)$ to the line is $\frac{y_0-mx_0-b}k$, so $7+3m-b=5k$ and $17-2m-b=10k$.
2. Subtracting, $10-5m=5k$, so $k=2-m$. Squaring, $1+m^2=4-4m+m^2$, so $m=\frac34$ and $k=\frac54$.
3. (If both centers were below the line, the same steps would give $k=m-2$, i.e. $m>2$, and squaring again forces $m=\frac34$, a contradiction. That second external tangent is the vertical line $x=-8$.)
4. Then $7+\frac94-b=\frac{25}4$, so $b=\boxed{3}$.