Answer
$(x-0)^2+(y-0)^2=\sqrt{13}$.
Work Step by Step
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: $r=\sqrt{(0-(-2))^2+(0-3)^2}=\sqrt{4+9}=\sqrt{13}.$
Also $C=(0,0)$, hence our equation: $(x-0)^2+(y-0)^2=\sqrt{13}$.
