Brian Lawrence showed me the following conceptual proof of Poncelet porism in the case of two circles, which I thought was neat and wanted to sketch here. (This is only a sketch, since I’m not really defining the integration.)

Let $P$ be a point on the outer circle, and let $Q$ be the point you get when you take the counterclockwise tangent from $P$ to the inner circle. Consider what happens if we nudge the point $P$ by a small increment $dP$.

The similar triangles in power of a point then give us the approximation

$$\frac{dP}{t(P)} = \frac{dQ}{t(Q)}$$

where $t(X)$ is the length of the tangent from $X$ on the large circle to the smaller one. (Note that because we’re working with circles, the definition of $t$ doesn’t care about clockwise vs counterclockwise tangent).

Now, suppose $(P_0, P_1, P_2, \dots, P_n)$ be a sequence of points on the large circle such that $P_i P_{i+1}$ is the counterclockwise tangent to the inner circle for all $i$. Now, suppose $(P_0', P_1', P_2', \dots, P_n')$ be another such sequence where $P_0'$ is slightly counterclockwise of $P_0$. Then we have the integral relations

$$\int_{P_0}^{P'_0} \frac{dX}{t(X)} = \int_{P_1}^{P'_1} \frac{dX}{t(X)} = \int_{P_2}^{P'_2} \frac{dX}{t(X)} = \dots = \int_{P_n}^{P'_n} \frac{dX}{t(X)}.$$

So if $P_0 = P_n$, it follows $P_0' = P_n'$ as well. Hence Poncelet’s closure theorem is proved.