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

Question

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)

Answer 1

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.

ORIGINAL ANSWER.

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 $r_1-r_2<AB<r_1+r_2$, 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 $AP-BP<r_1-r_2$ 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 $AB<r_1-r_2$, then by the triangular inequality $AP-BP<r_1-r_2$ 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.

Answer 2

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.