Answer:
[tex]\sf (x+2)^2+(y-2)^2=2[/tex]
Step-by-step explanation:
If RS is the diameter of the circle, then the midpoint of RS will be the center of the circle.
[tex]\sf midpoint=\left(\dfrac{x_s-x_r}{2}+x_r,\dfrac{y_s-y_r}{2}+y_r \right)[/tex]
[tex]\sf =\left(\dfrac{-1-(-3)}{2}+(-3),\dfrac{3-1}{2}+1 \right)[/tex]
[tex]\sf =(-2, 2)[/tex]
Equation of a circle: [tex]\sf (x-h)^2+(y-k)^2=r^2[/tex]
(where (h, k) is the center and r is the radius)
Substituting found center (-2, 2) into the equation of a circle:
[tex]\sf \implies (x-(-2))^2+(y-2)^2=r^2[/tex]
[tex]\sf \implies (x+2)^2+(y-2)^2=r^2[/tex]
To find [tex]\sf r^2[/tex], simply substitute one of the points into the equation and solve:
[tex]\sf \implies (-3+2)^2+(1-2)^2=r^2[/tex]
[tex]\sf \implies 1+1=r^2[/tex]
[tex]\sf \implies r^2=2[/tex]
Therefore, the equation of the circle is:
[tex]\sf (x+2)^2+(y-2)^2=2[/tex]
