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Prove that if 3 circles are tangent to one another then the tangents at the points of contact are concurrent.

I've tried using Ceva's Theorem here but it doesn't work since we don't know if the tangent passes through the center of the third circle if all the circles are of different sizes.

How else can I proceed?

Thanks.

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You can use a special line known as the radical axis.

The radical axis is the locus of all points on the plane that have the same power with respect to the 2 circles in question, i.e. the length of the tangent to both circles from any point on this line is the same.

It is know that for any 2 non-concentric (sharing a center) that are tangent to each other, the shared tangent is the radical axis.

Thus it can be shown that for any 3 circles, the 3 tangents formed will always be concurrent. The proof is as follows:

Let $w_1$,$w_2$ and $w_3$ be the 3 circles that are tangent to each other. We can assume that none of them share a center as it would be the same as 2 circles being tangent.

We know that for any point $P$ on the radical axis of $w_1$ and $w_2$, the power of P with respect to $w_1$ and $w_2$ is the same.

We also know that for any point $P$ on the radical axis of $w_2$ and $w_3$, the power of P with respect to $w_2$ and $w_3$ is the same.

Since the 2 lines must intersect (they cannot be parallel as this would imply that either they are not tangent to each other or at least 2 of them are identical and concentric), the point where the 2 radical axes meet has the same power with respect to $w_1$,$w_2$ and $w_3$. This implies that it must also be on the radical axis of $w_1$ and $w_3$. Therefore, this point is where the 3 tangents are concurrent.

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