Calculus: Early Transcendentals 8th Edition

$(x-3)^2 +(y+1)^2 = 25$
knowing the equation of the circle: $(x-h)^2 + (y-k)^2 = r^2$ we say that: $h= 3$ $k=-1$ $r=5$ then $(x-3)^2 + (y+1)^2 = 5^2$ $(x-3)^2 + (y+1)^2 = 25$