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