Answer
$\left(x-\frac{5}{2}\right)^2+(y-2)^2=\left(\frac{3}{2}\right)^2$.
$r=\frac{3}{2}$
$C=\left(\frac{5}{2},2\right)$
Work Step by Step
$C$ is the midpoint of the two given points, hence $C=\left(\frac{4+1}{2},\frac{2+2}{2}\right)=\left(\frac{5}{2},2\right).$
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: $r=\sqrt{\left(\frac{5}{2}-1\right)^2+(2-2)^2}=\sqrt{\frac{9}{4}+0}=\sqrt{\frac{9}{4}}=\frac{3}{2}.$
Hence our equation: $\left(x-\frac{5}{2}\right)^2+(y-2)^2=\left(\frac{3}{2}\right)^2$.
