Given two circles with centres $(x_1,y_1)$ and $(x_2,y_2)$ and radii $r_1$ and $r_2$ respectively. We get the following two equations.
$$ C_1 : (x-x_1)^2+(y-y_1)^2=r_1^2 $$ $$ C_2 : (x-x_2)^2+(y-y_2)^2=r_2^2 $$
Is there a way to determine the equations of the shared tangents of the two circles in general?
I have no idea where to even begin with such a question.
I like to think about it as a dilation in order to find the point where the shared tangents intersect. After that, you only need to calculate the intersections with the circles and find the slope.
An example:
Find the common external tangents of $(x-2)^2+(y-4)^2=15$ and $(x-4)^2+(y-8)^2=60$.
The example might be a special case because the tangents pass through the origin, but it works for any case. (However, translations can be very useful when solving this kind of problems.)
The circles can be obtained as a dilation from the other one ($k=2$, radius) and the centers are in $(2,4)$ and $(4,8)$, so: $$\overrightarrow{O}+\overrightarrow{V}=(2,4)\\\overrightarrow{O}+2\overrightarrow{V}=(4,8)$$ So the point where the external tangents intersect (which is also a point of the tangents) is $(0,0)$. If it weren't the origin, it would be very advisable to apply a translation so that points ends up in the origin.
Now we need a general equation of a line passing through $(0,0)$.$$y=mx$$This line intersects the first circumference once. Let's calculate the intersections.$$(x-2)^2+(mx-4)^2=15$$We obtain:$$(m^2+1)x^2-(8m+4)x+5=0$$If the line intersects the circumference once, that equation only has one solution $\implies$ the discriminant is 0. Which means:$$44m^2+64m-4=0$$That equation gives us the values of $m$. As I said before, a good idea is moving the intersection of the tangents to the origin. And for internal tangents you would use a negative dilation. If the radius is the same, the intersection of the tangents is a point at infinity, which means the slope of the line is the one of the line joining the centers of the circles.
