Answer
$(h,k)=(-\frac{1}{4},-1)$ and $r= 1$
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: $16x^{2}+16 y^{2}+8x +32y+1= 0$
The above equation can be written in the standard equation of the circle as follows:
$(x-(-\frac{1}{4}))^{2}+ (y-(-1))^{2} =1^{2}$
Compare it with equation (1), we have
Hence, center,$(h,k)=(-\frac{1}{4},-1)$ and $r= 1$
