Equals to $(x-5)^2+(y-5)^2=50$ $50$ can be written as sum of squares in two different ways as $(\pm 5)^2 + (\pm 5)^2$ and $(\pm 1)^2+(\pm 7)^2$ Conclusion follows

@above's sol is a lot cleaner than mine

$\frac{x^2 + y^2}{x+y}= 10$ $x^2+y^2=10x+10y$ $x^2-10x+y^2-10y=0$ $(x)^2-2\cdot x \cdot 5+(5)^2+(y)^2-2\cdot y \cdot5+(5)^2=5^2+5^2$ $(x-5)^2+(y-5)^2=50$ Now because $50=49+1=(\pm 7)^2+(\pm 1)^2$ and $50=25+25=(\pm 5)^2+(\pm 5)^2$ we have: case 1. $$(x-5)^2+(y-5)^2=(\pm 7)^2+(\pm 1)^2$$case 1.1. $$(x-5)^2=(7)^2, (y-5)^2=(1)^2\Rightarrow x=12,y=6$$case 1.2. $$(x-5)^2=(-7)^2, (y-5)^2=(-1)^2\Rightarrow x=-2,y=4$$case 1.3. $$(x-5)^2=(-7)^2, (y-5)^2=(1)^2\Rightarrow x=-2,y=6$$case 1.4. $$(x-5)^2=(7)^2, (y-5)^2=(-1)^2\Rightarrow x=12,y=4$$case 2. $$(x-5)^2+(y-5)^2=(\pm 5)^2+(\pm 5)^2$$case 2.1. $$(x-5)^2=(5)^2, (y-5)^2=(5)^2\Rightarrow x=y=10$$*case 2.2. $$(x-5)^2=(-5)^2, (y-5)^2=(-5)^2\Rightarrow x=y=0$$case 2.3. $$(x-5)^2=(-5)^2, (y-5)^2=(5)^2\Rightarrow x=0,y=10$$case 2.4. $$(x-5)^2=(5)^2, (y-5)^2=(-5)^2\Rightarrow x=10,y=0$$*Notice that in case 2.2. we have $x=y=0\Rightarrow x+y=0$ which we cannot have. Therefore, $(x,y)=(12,6),(-2,4),(-2,6),(12,4),(10,10),(0,10),(10,0)$ are the only solutions.

Rearrange the question to get $$(x-5)^2+(y-5)^2=50=(\pm1)^2+(\pm7)^2=(\pm5)^2+(\pm5)^2$$ $(x,y)=(12,6)(12,4)(10,10)(10,0)(0,10)(0,0)(-2,6)(-2,4)$ Note that $(x,y)=(0,0)$ is not a valid solution