## Precalculus (10th Edition)

$(x-2)^2+(y-2)^2=(\sqrt5)^2$.
$C$ is the midpoint of the two given points, hence $C=(\frac{4+0}{2},\frac{3+1}{2})=(2,2).$ 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: $r=\sqrt{(4-2)^2+(3-2)^2}=\sqrt{4+1}=\sqrt{5}.$ Hence our equation: $(x-2)^2+(y-2)^2=(\sqrt5)^2$.