<s>[INST] A horizontal line is drawn so that it is tangent to both circles centered at $A$ and $B$. It turns out that the radii drawn to the points of tangency are perpendicular to the line. First, prove that $AB$ must be the hypotenuse of a right triangle.
Use variables $r_1$ and $r_2$ to describe the radii. Prove that the distance $AB$ must be equal to the sum of the radii: $AB = r_1 + r_2$.
Use 2 right triangles to help you in your proof. Draw a secant line, $AB$. Draw a horizontal tangent to circle $B$ at the point of tangency, $P$.
Do the same for the first circle to find a point of tangency $R$ for that circle.  Draw the secant $AP$ and $BR$. Label $C$ the point of intersection when those two lines meet.

Because $A$, $B$, and the centers of both circles are not on the same horizontal line, the tangent to $A$ at $R$ must pass through the point $(h,k)$ below it at the base of $A$, and the tangent to $B$ at $P$ must pass through the point $(h',k')$ below it at the base of $B$. If $k=k'$, it follows that $AB = r_1 + r_2$.
Here's the figure:



For the above-diagram, what is the sign of $k$ (the height $AB$ is on above the vertical line through $P$)?[/INST]