# Line Joining Centers of Two Circles Touching Internally

## Theorem

Let two circles touch internally.

Then the straight line joining their centers passes through the point where they touch.

In the words of Euclid:

If two circles touch one another internally, and their centres, the straight line joining their centres, if it be also produced, will fall upon the point of contact of the circles.

## Proof

Let the circles $ABC$ and $ADE$ touch internally at $A$.

Let $F$ be the center of $ABC$ and let $G$ be the center of $ADE$.

We are to show that the straight line joining $F$ to $G$ passes through $A$.

Suppose, as in the diagram above, that it does not.

Let $FG$ fall on $H$ instead.

It will also pass through $D$ on its way, which lies on circle $ADE$.

Join $AF$ and $AG$.

From Sum of Two Sides of Triangle Greater than Third Side $AG + GF$ is greater than $AF$.

Therefore $AG + GF$ is greater than $FH$ as $F$ is the center and both $AF$ and $FH$ are radii.

So, subtract $FG$ from both $AG + GF$ and $FH$.

It follows that $AG$ is greater than $GH$, and hence also greater than $GD$.

But as $G$ is the center of $ADE$, it follows that $AG$ is supposed to be equal to $GD$.

So the $AG + GF$ can not be greater than $AF$ and so must be coincident with it.

Hence the result.

$\blacksquare$

## Historical Note

This proof is Proposition $11$ of Book $\text{III}$ of Euclid's The Elements.