Answer
$(x+1)^2+(y-5)^2=130$
Work Step by Step
RECALL:
The standard form of a circle whose center is at $(h, k)$ and radius $r$ is $(x--h)^2+(y-k)^2=r^2$.
The center is at $(-1, 5)$, then $h=-1$ and $k=5$.
Thus, the standard form of the given circle's equation of the circle is:
$[x-(1)]^2+(y-5)^2=r^2
\\(x+1)^2+(y-5)^2=r^2$
The circle passes through the point $(-4, -6)$.
This means that the coordinates of this point satisfy the equation of the circle.
Substitute the x and y values of the point into the equation to obtain:
$(x+1)^2+(y-5)^2=r^2
\\(-4+1)^2+(-6-5)^2=r^2
\\(-3)^2+(-11)^2=r^2
\\9+121=r^2
\\130=r^2$
Thus, the standard form of the given circle's equation is:
$(x+1)^2+(y-5)^2=130$
