Answer:
![\sf (x+2)^2+(y-2)^2=2](https://tex.z-dn.net/?f=%5Csf%20%28x%2B2%29%5E2%2B%28y-2%29%5E2%3D2)
Step-by-step explanation:
If RS is the diameter of the circle, then the midpoint of RS will be the center of the circle.
![\sf midpoint=\left(\dfrac{x_s-x_r}{2}+x_r,\dfrac{y_s-y_r}{2}+y_r \right)](https://tex.z-dn.net/?f=%5Csf%20midpoint%3D%5Cleft%28%5Cdfrac%7Bx_s-x_r%7D%7B2%7D%2Bx_r%2C%5Cdfrac%7By_s-y_r%7D%7B2%7D%2By_r%20%5Cright%29)
![\sf =\left(\dfrac{-1-(-3)}{2}+(-3),\dfrac{3-1}{2}+1 \right)](https://tex.z-dn.net/?f=%5Csf%20%3D%5Cleft%28%5Cdfrac%7B-1-%28-3%29%7D%7B2%7D%2B%28-3%29%2C%5Cdfrac%7B3-1%7D%7B2%7D%2B1%20%5Cright%29)
![\sf =(-2, 2)](https://tex.z-dn.net/?f=%5Csf%20%3D%28-2%2C%202%29)
Equation of a circle: ![\sf (x-h)^2+(y-k)^2=r^2](https://tex.z-dn.net/?f=%5Csf%20%28x-h%29%5E2%2B%28y-k%29%5E2%3Dr%5E2)
(where (h, k) is the center and r is the radius)
Substituting found center (-2, 2) into the equation of a circle:
![\sf \implies (x-(-2))^2+(y-2)^2=r^2](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20%28x-%28-2%29%29%5E2%2B%28y-2%29%5E2%3Dr%5E2)
![\sf \implies (x+2)^2+(y-2)^2=r^2](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20%28x%2B2%29%5E2%2B%28y-2%29%5E2%3Dr%5E2)
To find
, simply substitute one of the points into the equation and solve:
![\sf \implies (-3+2)^2+(1-2)^2=r^2](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20%28-3%2B2%29%5E2%2B%281-2%29%5E2%3Dr%5E2)
![\sf \implies 1+1=r^2](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%201%2B1%3Dr%5E2)
![\sf \implies r^2=2](https://tex.z-dn.net/?f=%5Csf%20%5Cimplies%20r%5E2%3D2)
Therefore, the equation of the circle is:
![\sf (x+2)^2+(y-2)^2=2](https://tex.z-dn.net/?f=%5Csf%20%28x%2B2%29%5E2%2B%28y-2%29%5E2%3D2)