Answer
$(x+1)^2+(y-3)^2=(\sqrt5)^2$.
Work Step by Step
$C$ is the midpoint of the two given points, hence $C=(\frac{1+(-3)}{2},\frac{4+2}{2})=(-1,3).$
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: $r=\sqrt{(-1-1)^2+(3-4)^2}=\sqrt{4+1}=\sqrt{5}.$
Hence our equation: $(x+1)^2+(y-3)^2=(\sqrt5)^2$.