A circular lamina of radius $x$ is removed from a circular lamina of radius $1$. If the center of gravity is at the edge of the smaller circle (along the line connecting the two centers) what is $x$?
I solved for the total area $A=\pi(1-x^2)$, divided by $2$ and set equal to the area of the segment $A=\frac{1}{2}(\theta-\sin(\theta))$. I can put $\theta$ in terms of $x$, but the resulting equation $\pi(1-x^2)=\theta-\sin(\theta)$ where $\theta=\arccos(8x^2-8x+1)$ seems to be impossible to solve without guessing and checking values for $x$.
I also tried to integrate turning the picture on its side, center the unit circle at the origin and calling the radius of the small circle "$a$" instead of "$x$". I then take an integral from $2a-1$ to $1$ of $\sqrt{1-x^2}$ and set equal to $\pi/4(1-a^2)$. Again I end up with an equation that has variables inside and outside of trig functions and I can't seem to solve for "$a$" without guessing and checking.
The attempts described above appear to be based on assuming that the center of gravity will be characterized by having equal areas (masses) on either side, but this is incorrect, because it is a weighted average of positions in your region. (For an example where the intuition might be clearer, imagine where the center of gravity is for a lamina consisting of $2$ circular regions with different radii, very far apart from one another.) Here the circular sector cut off by the chord, with area $\frac12(\theta-\sin(\theta)$, is on average closer to the center of gravity, which means it must have greater area than the other part.
A couple of previously unstated assumptions: (1) We deal with uniform density, so we are actually finding the centroid. (2) The smaller circle is tangent to the larger circle.
Let's center the circles on the positive $x$-axis at $(a,0)$ and $(1,0)$, so that they are tangent at the origin. This makes the region straightforward to describe in polar coordinates, as the circles have the equations $r=2a\cos(\theta)$ and $r=2\cos(\theta)$. We know the $y$-coordinate of the centroid is $0$, and the $x$-coordinate can be calculated as
$$\frac{\int\int_R x\,dA}{\mathrm{Area}(R)}.$$
Note that this is the weighted average referred to above, and you can find more general information about the definition and calculation of the centroid elsewhere (e.g. Wikipedia).
In polar coordinates the numerator becomes $$\int_{-\pi/2}^{\pi/2}\int_{2a\cos(\theta)}^{2\cos(\theta)}r^2\cos(\theta)\,dr\,d\theta=(1-a^3)\frac83\int_{-\pi/2}^{\pi/2}\cos^4(\theta)\,d\theta=\pi(1-a^3).$$
We divide by $\pi(1-a^2)$, set equal to $2a$, cancel the factor of $(1-a)$, and find that $a^2+a-1=0$. The relevant solution is $\dfrac{\sqrt5-1}{2}$.