Answer
$(x+8)^2+(y-1)^2=289$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the Center-Radius Form to find the equation of the circle with the following characteristics:
\begin{array}{l}\require{cancel}
\text{center }
(-8,1)
,\text{ passing through }
(0,16)
.\end{array}
$\bf{\text{Solution Details:}}$
With the given center, then $h=
-8
$ and $k=
1
.$
Using the Center-Radius Form of the equation of circles which is given by $(x-h)^2+(y-k)^2=r^2,$ then the equation of the circle is
\begin{array}{l}\require{cancel}
(x-(-8))^2+(y-1)^2=r^2
\\\\
(x+8)^2+(y-1)^2=r^2
\text{ (*)}
.\end{array}
Since the circle passes through the point $
(0,16)
,$ then substitute the coordinates in the equation. The equation above becomes
\begin{array}{l}\require{cancel}
(0+8)^2+(16-1)^2=r^2
\\\\
(8)^2+(15)^2=r^2
\\\\
64+225=r^2
\\\\
289=r^2
.\end{array}
Substituting $r^2$ in (*), the equation of the circle is
\begin{array}{l}\require{cancel}
(x+8)^2+(y-1)^2=289
.\end{array}