Say you are given a circle $C$ and a straight line $l$ exterior to the circle. How to describe the set of centers of circle that are tangent to both the $C$ and $l$?
I have no idea how to proceed. My first thought was to take the expression for the arbitrary circle and the line, and solve the systems that the intersections generate, then taking into account that since they are tangent, there can only be a single solution.
Since the equations are quadratic, this would then probably boil down to forcing the discriminant of a quadratic polynomial to be $0$ so that it only yields a single solution.
However I haven't been able to use this method, there are way too many variables and the equations just get huge and too messy. Any ideas?
Because of the content in which this exercise is proposed, I woulnd't be surprised if it were a conic section.
Let $A$ be the center of $C$, and $r$ be its radius. For any point $P$, let $|Pl|$ be the perpendicular distance from $P$ to $l$.
There are two ways a point $P$ can be on your locus. Either (1) the distance $|PA|=|Pl|+r$, or (2) $|PA|=|Pl|-r$. These two cases can be treated one by one.
For case (1), the condition $|PA|=|Pl|+r$ is the same as $|PA|=|Pm|$ where $m$ is a line parallel to $l$, but a distance of $r$ removed from $P$. Since $P$ must obviously be on the same side of $l$ as $A$ (this might be different if $l$ intersected $C$, but we assume that it doesn't), this is the same as saying that $m$ is a line parallel to $l$, but at a distance of $r$ farther away from $A$.
You should know the the locus of $P$ such that $|PA|=|Pm|$ is a parabola.
Case (2) can be handled similarly.