Answer
$\\\color{blue}{(x+8)^2+(y-1)^2=289}$
Work Step by Step
RECALL:
The center-radius form of a circle whose center is at $(h, k)$ and whose radius is $r$ units is:
$(x-h)^2+(y-k)^2=r^2$
The given circle has its center at $(-8, 1)$
Thus, the tentative equation of the circle is :
$(x-(-8))^2+(y-1)^2=r^2
\\(x+8)^2+(y-1)^2=r^2$
To find the value of $r^2$, substitute the x and y values of the given point on the circle $(0, 16)$ to obtain:
$(x+8)^2+(y-1)^2=r^2
\\(0+8)^2+(16-1)^2=r^2
\\8^2+15^2=r^2
\\64+225=r^2
\\289=r^2$
Therefore, the equation of the circle is:
$\\\color{blue}{(x+8)^2+(y-1)^2=289}$
