Answer
$(h,k)=(\frac{1}{4},-\frac{1}{4})$ and $r= (\frac{\sqrt {10}}{4})$
Work Step by Step
An equation of the circle with center $(h,k)$ and
radius $r$ is given as:
$(x-h)^{2}+(y-k)^{2}=r^{2}$ ...(1)
Given: $2x^{2}+2 y^{2}-x +y= 0$
The above equation can be written in the standard equation of the circle as follows:
$(x-\frac{1}{4})^{2}+ (y-(-\frac{1}{4}))^{2} =(\frac{\sqrt {10}}{4})^{2}$
Compare it with equation (1), we have
Hence, center,$(h,k)=(\frac{1}{4},-\frac{1}{4})$ and $r= (\frac{\sqrt {10}}{4})$
