#### Answer

$\color{blue}{x^2+(y-3)^2=40}$

#### 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 $(0, 3)$
Thus, the tentative equation of the circle is :
$(x-0)^2+(y-3)^2=r^2
\\x^2+(y-3)^2=r^2$
To find the value of $r^2$, substitute the x and y values of the point $(-2, 6)$ to obtain:
$(-2)^2+(3-6)^2=r^2
\\4+36=r^2
\\40=r^2$
Therefore, the equation of the circle is:
$\\\color{blue}{x^2+(y-3)^2=40}$