Question: Suppose I have the recurrence relation $a\_n = 3a\_{n-1} + 4a\_{n-2} -n +6,$ and I am given initial conditions $a\_0 = 0; a\_1 = 1$. I understand that without using initial conditions, the general solution $g(n,\alpha,\beta) = \alpha(4)^n +\beta(-1)^n + \frac{1}{6}n - \frac{25}{36} $ gives us a vector space for the $a\_n$ solution, and so any $g(n,\alpha\_1,\beta\_1) + g(n,\alpha\_2,\beta\_2)$ is also a solution.

But after making use of the initial conditions, I'm not sure how to prove (in a non-hand-waving way) that $a\_n = \frac{4^{n+1}}{9} + \frac{(-1)^n}{4} + \frac{1}{6}n - \frac{25}{36}$ is the unique solution for the recurrence relation.

Answer: 
The given circle has a radius of $2$ and the center of $(1,1)$

The distance from the center of the given circle to the midpoint of cords is a constant of $1$ due to the fact that in the right triangle formed by the center, the midpoint and one end of the cord we have a $30$ degree angle opposite to the segment connecting the center to the midpoint.

Therefore the locus is a circle with the same center and the radius $1$ that is $$(x-1)^2 + (y-1)^2 =1$$
Clearly all such chords are the same length.

Any rotation around the origin that sends one chord to another must send the midpoint of the first to the second, so we can obtain all midpoints by rotating one around the centre, implying the locus is a circle with the same centre as the original circle.

We now need to find one midpoint to identify the radius. Choosing a vertical chord here makes calculation easier.