Circles with centers $(2,4)$ and $(14,9)$ have radii $4$ and $9$, 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 $b$? $\mathrm{(A) \ } \frac{908}{119}\qquad$ $ \mathrm{(B) \ } \frac{909}{119}\qquad$ $ \mathrm{(C) \ } \frac{130}{17}\qquad$ $ \mathrm{(D) \ } \frac{911}{119}\qquad$ $ \mathrm{(E) \ } \frac{912}{119}$
Solution
For an external tangent $y=mx+b$ (i.e. $mx-y+b=0$), both centers lie on the same side, so $$\frac{2m-4+b}4=\frac{14m-9+b}9.$$
Cross-multiplying: $9(2m-4+b)=4(14m-9+b)$, giving $5b=38m$, so $b=\frac{38m}5$.
Substituting into the distance condition $|2m-4+b|=4\sqrt{m^2+1}$: $\left|\frac{48m-20}5\right|=4\sqrt{m^2+1}$, so $|48m-20|=20\sqrt{m^2+1}$.
Squaring: $2304m^2-1920m+400=400m^2+400$, so $1904m^2=1920m$, giving $m=\frac{120}{119}$.
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 $b$?
$\mathrm{(A) \ } \frac{908}{119}\qquad$ $ \mathrm{(B) \ } \frac{909}{119}\qquad$ $ \mathrm{(C) \ } \frac{130}{17}\qquad$ $ \mathrm{(D) \ } \frac{911}{119}\qquad$ $ \mathrm{(E) \ } \frac{912}{119}$ 