Algebra

$2(x^2+y^2)=x+y+8\\ 2(x^2-\frac{1}{2}x+y^2-\frac{1}{2}y)=8\\ x^2-\frac{1}{2}x+y^2-\frac{1}{2}y=4\\ x^2-\frac{1}{2}x+\frac{1}{16}+y^2-\frac{1}{2}x+\frac{1}{16}=4+\frac{1}{8}\\ (x-\frac{1}{4})^2+(y-\frac{1}{4})^2=\frac{33}{8}$

Notice that for the equation $x^2+y^2=1$, we can substitute $x=\cos(\theta)$ and $y=\sin(\theta)$.

Similarly, here we can substitute $x=\frac{\sqrt{33}}{2\sqrt{2}}(\cos(\theta))+\frac{1}{4}, y=\frac{\sqrt{33}}{2\sqrt{2}}(\sin(\theta))+\frac{1}{4}$, but with slight adjustments due to the translation and the dilation of the circle.

So, we want to find the maximum value of:

$(\frac{\sqrt{33}}{2\sqrt{2}}(\cos(\theta))+\frac{1}{4})-(\frac{\sqrt{33}}{2\sqrt{2}}(\sin(\theta))+\frac{1}{4})\\ =\frac{\sqrt{33}}{2\sqrt{2}}(\cos(\theta)-\sin(\theta))$

I know there is a way to find a simpler form of $\cos(\theta)-\sin(\theta)$, but it's hard to see and I didn't see it, so what I ended up doing is just taking the derivative with respect to theta, which is:

$-\sin(\theta)-\cos(\theta)$

The extreme points are when:

$-\sin(\theta)=\cos(\theta)$, which only happens when $\theta = 135+180n$, where n is an integer. From now, It shouldn't be hard to figure out that the maximum value of $\cos(\theta)-\sin(\theta)$ is $\sqrt{2}$.

This means that the maximum value of our original expression is $\frac{\sqrt{33}}{2\sqrt{2}} \cdot \sqrt{2}=\boxed{\frac{\sqrt{33}}{2}}$

For completeness, try to find which values of x and y give that maximum value.

Edit: I noticed that my previous answer contained a huge mistake lol

Hi Textot,

What you have done looks very impressive.   

I have to think about it some more.