$ I(r) = \int_0^{2\pi}\frac{\cos(t) - r}{1 - 2r\cos t + r^2}\,dt$ is always zero for $r\in[0,1)$. Why?

An alternative proof using complex methods:

For $0< r < 1$ let $$ f_r \colon B_\frac{1}{r} (0) \to \mathbb{C} \, , \, f_r(z) = \frac{- \ln(1-rz)}{z} \, , $$ where $f_r(0) = r $ . Then $f_r$ is holomorphic, so $$ I(r) \equiv - \int \limits_0^{2\pi} \ln(1-r \mathrm{e}^{\mathrm{i}t}) \, \mathrm{d} t = - \mathrm{i} \int \limits_{S^1} f_r(z) \, \mathrm{d} z = 0 $$ holds by Cauchy's theorem. If you are not familiar with complex analysis, you can also show this using the Taylor series of the logarithm: $$ I(r) = \sum \limits_{n=1}^\infty \frac{r^n}{n} \int \limits_0^{2\pi} \mathrm{e}^{\mathrm{i} n t} \, \mathrm{d} t = 0 \, . $$ This implies \begin{align} \int \limits_0^{2\pi} \frac{\cos(t) - r}{1 - 2 r \cos(t) + r^2} \, \mathrm{d} t &= - \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} r} \int \limits_0^{2\pi} \ln(1 - 2 r \cos(t) + r^2) \, \mathrm{d} t \\ &= - \frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} r} \int \limits_0^{2\pi} \ln[(1 -r \mathrm{e}^{\mathrm{i} t})(1 -r \mathrm{e}^{-\mathrm{i} t})] \, \mathrm{d} t \\ &= \frac{\mathrm{d}}{\mathrm{d} r} I(r) = \frac{\mathrm{d}}{\mathrm{d} r} 0 = 0 \end{align} as desired.

(Since it is not mentioned yet) The function $$ P_r(t) = \sum_{n=-\infty}^\infty r^{|n|} e^{int} =\frac{1-r^2}{1-2r\cos t +t^2},\quad0\le r<1, 0\le t\le 2\pi $$ is called the Poisson kernel. For any continuous function $f:[0,2\pi]\to \Bbb C$ the Poisson integral $$ u(r,\theta) = \frac{1}{2\pi i}\int_0^{2\pi} f(t)P_r(\theta-t)\mathrm{d}t\tag{*} $$ gives the unique solution of the Dirichlet problem $\triangle u =0, \lim\limits_{r\to 1^-}u(r,\theta)=f(\theta)$.

Let $u(r,\theta)=r\cos \theta$. Since it is a real part of the analytic function $z=r e^{i\theta}$, we find that $$\triangle u=0,\ \ \ \lim\limits_{r\to 1^-}u(r,\theta)=\cos \theta.$$ By $\text{(*)}$ it follows that $$ r\cos \theta =\frac{1}{2\pi}\int_0^{2\pi} \frac{\cos t(1-r^2)}{1-2r\cos (t-\theta)+r^2}\mathrm{d}t $$ for all $0\le r<1$ and $0\le \theta\le 2\pi$. Using the fact $$ \frac{1}{2\pi}\int_0^{2\pi} \frac{1-r^2}{1-2r\cos (t-\theta)+r^2}\mathrm{d}t=\frac{1}{2\pi}\sum_{n=-\infty}^\infty r^{|n|} \int_0^{2\pi}e^{in(t-\theta)}\mathrm{d}t =1, $$ it follows $$ \frac{1}{2\pi}\int_0^{2\pi} \frac{\left(\cos t-r\cos \theta\right)(1-r^2)}{1-2r\cos (t-\theta)+r^2}\mathrm{d}t =0. $$ By letting $\theta =0$, we obtain $$ \frac{1-r^2}{2\pi }\int_0^{2\pi} \frac{\cos t-r}{1-2r\cos t+r^2}\mathrm{d}t =0,$$ or equivalently $$ \int_0^{2\pi} \frac{\cos t-r}{1-2r\cos t+r^2}\mathrm{d}t =0.$$

It seems to me that this result is a particular case (or rather 1D adaptation) of Newton's shell theorem. In 2D, the theorem states that no net gravitational force is perceived by objects located inside a uniform hollow sphere.

Here, consider the integral $$ \vec I(M) = \int_{X\in C}\frac{\vec{MX}}{MX^2}\mathrm ds $$ where $M$ is a point at a distance $r<1$ away from the center $O$ of the unit circle $C$, and $\mathrm ds$ is a length element of the circle. What you are asking for is a component of $\vec I$, the one directed along $\vec{OM}$. In fact, we have $\vec I=\vec 0$. There is a nice geometric proof of this in 2D due to Newton; see here. You can try to adapt it to 1D.