Given two circles outside each other, the maximum number of common tangents to the two circles is 4 according to Wikipedia. How can this be proven? That's a picture above showing what I'm talking about. How can we prove that no other common tangents can be drawn aside those four?

Why the downvote? – Toba Dec 10 '20 at 11:05

I am neither the downvoter nor the closevoter. But some context would be nice, for example a picture showing that $4$ is possible at all. – Peter Dec 10 '20 at 11:10

I find the downvote quite painful though as this is the first question I'm asking here. I would've added an image if I had more reputation points. – Toba Dec 10 '20 at 11:24

2Don't take downvotes and closevotes too serious as long as you have a positive score and the question is not closed. – Peter Dec 10 '20 at 11:51

1In your title "distinct" is not the appropriate adjective: you should say "exterior one to the other" – Jean Marie Dec 11 '20 at 01:29
4 Answers
Hint Let the equation of the tangent be $a x + b y = c$ with $a^2+b^2=1$, and let $(x_i, y_i), r_i$ be the centers and radius of the circles. The following system must hold \begin{equation} \begin{array}\cr a x_1 + b y_1  c = \pm r_1\cr a x_2 + b y_2  c = \pm r_2\cr a^2 + b^2 = 1\cr \end{array} \end{equation} The two first equations with unknowns $(a, b, c)$ form a linear system with rank 2 if the centers are different, hence for each of the 4 right hand sides, there is a straight line of solutions $(a, b, c)$. Two points at most on this straight line satisfy $a^2 + b^2 = 1$. It gives at most 8 solutions but these solutions come in identical pairs because $(a, b, c)$ defines the same line as $(a, b, c)$, hence there are at most 4 tangent lines.
Where does the system come from ? The equation of a line $L$ in the plane can be written $a x + b y = c$ where $(a, b)\not = (0,0)$. Multiplying $a, b, c$ by a non zero factor does not change the line defined by this equation, so we can as well multiply by $1/\sqrt{a^2+b^2}$ to obtain an equation where finally $a^2 + b^2=1$. If $P = (x_i, y_i)$ is any point in the plane, the distance between $P$ and the line $L$ is \begin{equation} d(P, L) = a x_i + b y_i  c \end{equation} Indeed for any $t\in {\mathbb R}$, the point $(x_i + t a, y_i + t b)$ belongs to the line perpendicular to $L$ passing through $P$, because the vector $(a, b)$ is orthogonal to $L$. This point belongs to $L$ iff \begin{equation} a (x_i + t a) + b(y_i+ t b) = c \end{equation} which implies $t = (a x_i + b y_i  c)$. The distance from $P$ to the intersection point is then $t = a x_i + b y_i  c$.
Thus, the two first equations of the system simply say that the distance of the center of each circle to the line $L$ is $r_i$. This property characterizes the tangent lines of the circles.
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I have very little knowledge of linear algebra, making this answer difficult for me to understand. I'll really appreciate an answer using analytic geometry and the equations of the circles. – Toba Dec 10 '20 at 11:39



1@Toba Without the linear algebra terms, I might say: "For each of the four choices of signs, the two linear equations together have three unknowns. There is 1 "unknown"/degree of freedom left (assuming the equations are actually different). The quadratic equation pins down at most 2 solutions, for each choice of sign. Thus, there are at most 8 solutions for $(a,b,c)$. Since the solutions come in pairs of opposite sign, there are up to 4 different tangent lines possible." The linear algebra is only there to justify thinking "2 equations + 3 unknowns = 1 unknown". – HTNW Dec 10 '20 at 19:24

@Gribouillis can you be more explicit in your explanation of how the three equations were gotten? – Toba Dec 13 '20 at 16:15

This approach is by geometric construction.
Given circles exterior to one another with centers $A$, $B$: join $AB$, cutting the circles at $D$, $E$, and extend $AB$ to $C$ such that$$\frac{AC}{BC}=\frac{AD}{BE}$$From $C$ draw tangent $CF$ and join $BF$. Draw $AG\parallel BF$.
It is clear that $AF$ extended makes$$\triangle CBF\sim\triangle CAG$$so that $\angle AGC$ is right, and $CFG$ is a common tangent to both circles. Next, take $H$ on $DE$ such that$$\frac{HD}{HE}=\frac{AD}{BE}$$Draw tangent $HJ$ and join $BJ$. Make $AK\parallel BJ$. Again it is clear that $JH$ extended makes$$\triangle HJB\sim\triangle HKA$$so that $\angle HKA$ is right and $JHK$ is tangent to both circles.
Since each of these two tangents has its symmetrical counterpart, we then have four common tangents. Further, except in the special case of equal radii, a common tangent must intersect $AB$ or its extension. And since $C$ on $AB$ outside the circles, and $H$ on $DE$ between the circles, are unique points, then only four common tangents are possible.
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I believe this can be made more rigorous, but here is an intuitive argument. Imagine a tangent to one of the circles meeting it at a point. As you rotate that circle once around its centre the two ends of the tangent will each pass through the other circle. Each end of the tangent will have a point of contact as it enters the other circle and as it leaves.
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Any common tangent line must fall in one of the the following two cases: the centers of the circles, $A$ and $B,$ are on the same side of the tangent line, or they are on opposite sides of the tangent line.
When the centers are on the same side, first consider the case where the line connecting the centers of the circles intersects the tangent line at $C$. The point $C$ cannot be between $A$ and $B$ (because then the centers would be on opposite sides of the tangent line); without loss of generality, suppose the points $A,$ $B,$ and $C$ occur in that order along their common line. Since the tangent is perpendicular to the radius at the tangent point, we have two similar right triangles, one with hypotenuse $AC$ and one with hypotenuse $BC,$ where the ratio $AC:BC$ is the ratio of the radii of the circles.
Using this ratio and the order of the points on their common line, we can construct the location of the point $C$ uniquely. The point of tangency of the circle $B$ with the tangent through $C$ is the intersection of circle $B$ with the circle constructed on the diameter $BC.$ There are exactly two such intersections, so we can construct exactly two tangent lines in this case. We observe that the radii are unequal in this case.
In the case where the centers are on the same side of the tangent and the tangent does not intersect the line between the centers, the tangent is parallel to the line between the centers. The radius from $B$ to the tangent point is perpendicular to the line between the centers; so we construct a perpendicular to $AB$ through $B$, which intersects the circle $B$ at two points, which produce the two possible tangent lines in this case. We observe that the radii are equal in this case, so the two cases where the centers are on the same side of the tangent are mutually exclusive; we get only two tangent lines from that general case.
In the case where the centers are on opposite sides of the tangent line, the line connecting the centers must intersect the tangent line between the centers. Let $D$ be that intersection point. Again we have two similar right triangles, one with vertices at $A$, $D$, and the point of tangency on circle $A$, the other with vertices at $B$, $D$, and the point of tangency on circle $B$, so $AD:BD$ is the ratio of the two radii. This lets us uniquely construct the position of $D.$ The point of tangency with circle $B$ is then the intersection of that circle with the circle constructed on the diameter $BD,$ which gives two possible points of tangency and two possible tangent lines.
So we have at most two tangent lines that have both centers on the same side and at most two tangent lines that have the centers on opposite sides. The arguments above do not show that either set of tangent lines exist; to do that, we would have to consider whether each circle is inside, outside, or intersecting the other. But if you accept the figure in the question as demonstrating the existence of four tangent lines in the case where the circles are mutually outside each other, we have shown there are no other tangents.
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