A circle of radius $2$ has center at $(2,0)$. A circle of radius $1$ has center at $(5,0)$.  A line is tangent to the two circles at points in the first quadrant.  What is the $y$-intercept of the line?
Let's call the first circle circle A, and the second circle circle B. Also, let's call the point of tangency of the circle with radius $2$ point $B$. Let's call the point where the line intersects the $y$-axis point $C$. Let's call the center of the circle with radius $1$ point $D$. In order for the line to be tangent to the circle with radius $2$ at point $B$, angle $ABD$ must be a right angle. That means that the line is perpendicular to $BD$. BD is a radius of the circle with radius $2$, so it is $2$ units long. AE is a radius of the circle with radius $1$, so it is $1$ unit long. Right.  Notice that $AB$ is the hypotenuse of right triangle $ABD$. Right.  So triangle $FAE$ is a right triangle. So triangle $FBD$ is a right triangle. That's true.  And that means that triangle $FBD$ is similar to triangle $FAE$ Right.  So, we have $\frac{FB}{FA} = \frac{BD}{AE} = \frac{2}{1}$ That's right. So, $FB = 2FA$. Right.  So, $FD = 2FE$. Substituting, we have $2FE = DE + EF$, so $DE = EF$. So $EF = DE$ The distance between the centers of the given circles is 3. So, $\sin F = \frac{opposite}{hypotenuse} = \frac{1}{3}$ So, triangle $\triangle FCO$ is a right triangle And $OC$ is the y-intercept we're looking for. So, $OC = OF \tan \angle F = 8 \tan \angle F = 8 \tan \sin^{-1} \frac{1}{3}$ And since $\tan \sin^{-1} \frac{1}{3} = \frac{\sqrt{8}}{3}$, we have $OC = 2\sqrt{2}$

