Answer
(a) center-radius form: $\color{blue}{(x-3)^2+(y-1)^2=4}$
(b) general form: $\color{red}{x^2+y^2-6x-2y+6=0}$
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 $(3, -1)$ and $(3, 3)$. Note that the midpoint of these two points is $(3, 1)$.
Thus, the circle has its center at $(3, 1)$.
The radius of the circle can be found by finding the distance from the center $(3, 1)$ to $(3, 3)$, which is 2 units.
Thus, the radius of the circle is $2$.
(a) The circle has its center at $(3, 1)$ and has a radius of $2$ units.
Therefore, the equation of the circle in center-radius form is:
$(x-3)^2+(y-1)^2=2^2
\\\color{blue}{(x-3)^2+(y-1)^2=4}$
(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:
$(x-3)^2+(y-1)^2=4
\\x^2-6x+9+y^2-2y+1=4
\\x^2+y^2-6x-2y+10=4
\\x^2+y^2-6x-2y+10-4=0
\\\color{red}{x^2+y^2-6x-2y+6=0}$
