Answer
center,$(h,k)=(2,-5)$ and $r=4$
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: $x^{2}+ y^{2}-4x +10y +13 = 0$
The above equation can be written in the standard equation of the circle as follows:
$(x-2)^{2}+ (y-(-5))^{2} = 4^{2}$
Compare it with equation (1), we have
Hence, center,$(h,k)=(2,-5)$ and $r=4$
