Answer
$(x-\frac{1}{4})^2+(y-\frac{1}{4})^2=4$
Center: $(\frac{1}{4},\frac{1}{4})$
Radius: $r=2$
Work Step by Step
$2x^2+2y^2-2x-2y-7=0$ (divide the equation by 2):
$x^2+y^2-x-y-\frac{7}{2}=0$
Notice that $-x=-2(\frac{1}{2})x$ and that $(\frac{1}{2})^2=\frac{1}{4}$
Notice that $-y=-2(\frac{1}{2})y$ and that $(\frac{1}{2})^2=\frac{1}{4}$
$x^2-x+\frac{1}{4}-\frac{1}{4}+y^2-y+\frac{1}{4}-\frac{1}{4}-\frac{7}{2}=0$
$(x^2-x+\frac{1}{4})+(y^2-y+\frac{1}{4})=\frac{7}{2}+\frac{1}{4}+\frac{1}{4}$
$(x-\frac{1}{4})^2+(y-\frac{1}{4})^2=\frac{7}{2}+\frac{2}{4}=\frac{7}{2}+\frac{1}{2}=4$
$(x-\frac{1}{4})^2+(y-\frac{1}{4})^2=4$
The equation of a circle in standard form:
$(x-h)^2+(y-k)^2=r^2$ in which $(h,k)$ is the center and $r$ is the radius
$(x-\frac{1}{4})^2+(y-\frac{1}{4})^2=2^2$
Center: $(\frac{1}{4},\frac{1}{4})$
Radius: $r=2$
