Answer
center,$(h,k)=(-\frac{1}{2},0)$ and $r= \frac{1}{2}$
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: $x^{2}+ y^{2}+x = 0$
The above equation can be written in the standard equation of the circle as follows:
$(x-(-\frac{1}{2}))^{2}+ (y-0)^{2} = (\frac{1}{2})^{2}$
Compare it with equation (1), we have
Hence, center,$(h,k)=(-\frac{1}{2},0)$ and $r= \frac{1}{2}$
