For a circle w/ a unit radius, how far from the center of that circle must the center of a chord be for the area under that chord (the one that would not include the circle's center) to be A?
If $theta$ is the central angle of the chord in radians, then
$theta$ - sin $theta$ = 2A
but I don't see how to get $theta$ from that. (Once I have $theta$, I can figure out the distance).
If there is no analytical solution to the general case, is there one for a specific case, say A = $\pi/4$?
On
If you put the origin at the center of the circle, then the area under the cord is $$ A=\int _{-1}^{\frac {1}{2}-d} 2\sqrt {1-y^2}dy$$
While finding the integral is not too hard solving the result for $d$ is possible only with numerical methods.
On
Staying with your equation $$\frac{1}{2} (\theta -\sin (\theta ))=A$$ which, as already said in answers, does not show analytical solution, we can however notice that the series expansion of the rhs built at $\theta=0$ $$\frac{1}{2} (\theta -\sin (\theta ))=\frac{\theta ^3}{12}-\frac{\theta ^5}{240}+\frac{\theta ^7}{10080}-\frac{\theta ^9}{725760}+\frac{\theta ^{11}}{79833600}+O\left(\theta ^{13}\right)$$ almost perfectly matches for $0 \leq \theta \leq \frac \pi 2$ and is more than acceptable up to $\theta=\pi$.
So, for a try, let us use series reversion to get $$\theta=x+\frac{x^3}{60}+\frac{x^5}{1400}+\frac{x^7}{25200}+\frac{43 x^9}{17248000}+O\left(x^{11}\right)\qquad \text{where} \qquad x=\sqrt[3]{12A}$$
In order to check how good or bad this works, give $\theta$ a value in order to compute $A$ and recompute $\theta$ using the above truncated series.
Below are give some results
$$\left( \begin{array}{cc} \theta_{given} & \theta_{calc} \\ 0.0 & 0.00000 \\ 0.1 & 0.10000 \\ 0.2 & 0.20000 \\ 0.3 & 0.30000 \\ 0.4 & 0.40000 \\ 0.5 & 0.50000 \\ 0.6 & 0.60000\\ 0.7 & 0.70000 \\ 0.8 & 0.80000 \\ 0.9 & 0.90000 \\ 1.0 & 1.00000 \\ 1.1 & 1.10000 \\ 1.2 & 1.20000 \\ 1.3 & 1.30000 \\ 1.4 & 1.39999 \\ 1.5 & 1.49999 \\ 1.6 & 1.59998 \\ 1.7 & 1.69996 \\ 1.8 & 1.79992 \\ 1.9 & 1.89987 \\ 2.0 & 1.99978 \\ 2.1 & 2.09964 \\ 2.2 & 2.19943 \\ 2.3 & 2.29911 \\ 2.4 & 2.39866 \\ 2.5 & 2.49801 \\ 2.6 & 2.59710 \\ 2.7 & 2.69585 \\ 2.8 & 2.79418 \\ 2.9 & 2.89196 \\ 3.0 & 2.98906 \end{array} \right)$$
If you want a more comact form, we could build from the series the Padé approximant $$\theta=x \frac {1-\frac{1493 }{21120}x^2+\frac{167 }{268800}x^4 } {1-\frac{123 }{1408}x^2+\frac{403 }{295680}x^4 }\qquad \text{where} \qquad x=\sqrt[3]{12A}$$
Edit
Strating from Yves Daoust's answer $$A=2\arccos(d)- d\sqrt{1-d^2}$$ the same process would lead to the approximation $$d=x+\frac{x^3}{18}+\frac{x^5}{1080}-\frac{1439 x^7}{136080}-\frac{374399 x^9}{29393280}-\frac{115951679 x^{11}}{9699782400}+O\left(x^{13}\right)$$ where $x=\frac{\pi-A} 3$
For an aperture $\theta=2\alpha$, the area of the segment is $2\alpha-\cos\alpha\sin\alpha$ and the distance to the chord is $\cos\alpha$. In terms of the distance $d$, the area is
$$A=2\arccos(d)- d\sqrt{1-d^2}$$
Unfortunately, there is no analytical inverse and you need to resort to a numerical method.