Answer
(a) center-radius form: $\color{blue}{(x+1)^2+(y+2)^2=9}$
(b) general form: $\color{red}{x^2+y^2+2x+4y-4=0}$
Work Step by Step
RECALL:
(1) The center-radius form of a circle's equation is $(x-h)^2+(y-k)^2=r^2$ with center at $(h, k)$ and a radius of $r$ units.
(2) The general form of a circle's equation is $x^2+y^2+Dx+Ey+F=0$
The center of the circle can be found by finding the the midpoint of $(-1, 1)$ and $(-1, -5)$. Note that the midpoint of these two points is $(-1, -2)$.
Thus, the circle has its center at $(-1, -2)$.
The radius of the circle can be found by finding the distance from the center $(-1, -5)$ to $(-1, -2)$, which is 3 units.
Thus, the radius of the circle is $3$.
(a) The circle has its center at $(-1, -2)$ and has a radius of $3$ units.
Therefore, the equation of the circle in center-radius form is:
$(x-(-1))^2+(y-(-2))^2=3^2
\\\color{blue}{(x+1)^2+(y+2)^2=9}$
(b) To find the equation of the circle in general form, expand the equation in (a) above then put all terms on the left side to obtain:
$(x+1)^2+(y+2)^2=9
\\x^2+2x+1+y^2+4y+4=9
\\x^2+y^2+2x+4y+5=9
\\x^2+y^2+2x+4y+5-9=0
\\\color{red}{x^2+y^2+2x+4y-4=0}$
