# How to prove that two tangent circles have a common tangent at their point of tangency?

I tried to prove that the centers of two tangent circles are colinear with the point of tangency. If I just assume they have a common tangent at their point of tangency then it's pretty easy. However, how do we prove that two circles do indeed have a common tangent at their point of tangency? (without using the collinearity of the centers and the point of tangency to avoid circular logic)

• Having a common tangent line at the intersection point is more or less the definition of two curves being tangent at that point. How is tangency defined in your context? Jul 15 at 10:06
• @Goblin Claus. Welcome to MSE. Consider two circles intersect at two points. connect these points. The mid point of this segment is the one two points of intersection join when the circles are tangent. Jul 15 at 10:13
• The definition that is given is : "two circles are tangent if they have only one point of intersection" Jul 15 at 10:15
• Don’t you think that this problem ultimately boils down to proving the collinearity of the two centers and the point of tangency of two circles? That is exactly what @Intelligentipauca has done in his edited answer. If you want I can give another proof of that.
– YNK
Jul 16 at 8:59
• Yeah. I agree. If we have the theorem about collinearity proven independently of the existence of a common tangent then it's fine. Moreover, it's then quite easy to prove the existence of a common tangent at the point of tangency by showing the angle between the two "different" tangents to the circles is actually 0. Jul 16 at 10:24

EDIT. (Original answer at the end).

The detailed analysis below is not really necessary: you can reason by contradiction.

Suppose two circles, with centres $$A$$ and $$B$$, are tangent at a point $$P$$ NOT lying on line $$AB$$. Construct point $$P'$$, reflection of $$P$$ about line $$AB$$. Then $$P'\ne P$$ and we have $$P'A=PA$$, $$P'B=PB$$. Hence $$P'$$ lies on both circles and is a second point of intersection between them. But that is impossible, because the circles are tangent and must have a single intersection point, QED.

Let $$A$$, $$B$$ be the centres of the circles, $$r_1$$, $$r_2$$ their radii, with $$r_1\ge r_2$$. You must consider five possible cases.

1. If $$AB>r_1+r_2$$, then by the triangular inequality $$AP+BP>r_1+r_2$$ for any point $$P$$ in the plane. Hence the circles have no point in common.

2. If $$AB=r_1+r_2$$, then by the triangular inequality $$AP+BP>r_1+r_2$$ if $$P\notin AB$$ and $$AP+BP=r_1+r_2$$ for any point $$P\in AB$$. In particular, there exists on $$AB$$ a unique point $$P$$ such that $$AP=r_1$$, and consequently $$BP=r_2$$ Hence the circles have a single point in common: they are externally tangent.

3. If $$r_1-r_2, then a triangle with sides $$r_1$$, $$r_2$$ and $$d=AB$$ can exist. Construct on ray $$AB$$ point $$H$$ such that $$AH={r_1^2-r_2^2+d^2\over2d}$$. On the line through $$H$$, perpendicular to $$AB$$, we can then construct two points $$P$$ and $$P'$$ such that $$HP=HP'=\sqrt{r_1^2-AH^2}.$$ You can check that $$AP=AP'=r_1$$ and $$BP=BP'=r_2$$. Hence the circles have (at least) two points in common.

4. If $$AB=r_1-r_2$$, then by the triangular inequality $$AP-BP if $$P$$ doesn't lie on ray $$AB$$, hence $$P$$ cannot lie on both circles. On the other hand, there exists a unique point $$P$$ on ray $$AB$$ such that $$AP=r_1$$, and consequently $$BP=r_2$$. Hence the circles have a single point in common: they are internally tangent.

5. If $$AB, then by the triangular inequality $$AP-BP for any point $$P$$ on the plane, hence $$P$$ cannot lie on both circles.

As you can see, only in cases 2 and 4 the circles are tangent. In both cases tangency point $$P$$ lies on line $$AB$$, QED.

• If AB=d<r1+r2 , then there exists on AB a unique point H such that AH=(r1)^2-(r2)^2+d^2/2d can you explain this one please? Jul 15 at 20:04
• @GoblinClaus You are right: there are more cases to consider. I'll think of it. Jul 15 at 20:06
• I found a much simpler reasoning, see my edit. Jul 16 at 6:38
• Bellissimo! Thanks a lot! Jul 16 at 10:32

Hint.

Suppose we have two tangent circles:

$$\cases{ C_1 \to (x-1)^2+(y-1)^2 = \frac 12\\ C_2 \to (x-2)^2+(y-2)^2 = \frac 12\\ }$$

Eliminating $$y$$ between then we have

$$4x^2-12x+9=(3-2 x)^2=0$$

with a double root at $$x=\frac 32$$ evidencing its tangency. Those two points are common to $$C_1$$ and $$C_2$$ hence the tangent is common.