Answer
$r=\sqrt2$
$(x-1)^2+(y-2)^2=(\sqrt2)^2$.
$C=(1,2)$
Work Step by Step
$C$ is the midpoint of the two given points, hence $C=(\frac{0+2}{2},\frac{1+3}{2})=(1,2).$
The general equation for a circle with radius $r$ and centre $(h,k)$ is: $(x-h)^2+(y-k)^2=r^2$.
The distance formula from $P_1(x_1,y_1)$ to $P_2(x_2,y_2)$ is $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.
Hence here: $r=\sqrt{(2-1)^2+(3-2)^2}=\sqrt{1+1}=\sqrt{2}.$
Hence our equation: $(x-1)^2+(y-2)^2=(\sqrt2)^2$.
