Line Joining Centers of Two Circles Touching Internally

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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.

(The Elements: Book $\text{III}$: Proposition $11$)


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.


Historical Note

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