Answer:
c) [tex]\sf r^2=(x+5)^2+(y-4)^2[/tex]
Step-by-step explanation:
The center of the circle will be the midpoint of the endpoints.
[tex]\sf midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)[/tex]
Given:
[tex]\sf \implies midpoint=center=\left(\dfrac{-8-2}{2},\dfrac{2+6}{2}\right)=(-5,4)[/tex]
Equation of a circle: [tex]\sf (x-a)^2+(y-b)^2=r^2[/tex]
(where (a, b) is the center and r is the radius)
Substituting center (-5, 4) into the equation:
[tex]\sf \implies (x-(-5))^2+(y-4)^2=r^2[/tex]
[tex]\sf \implies (x+5)^2+(y-4)^2=r^2[/tex]
Therefore the solution is C: [tex]\sf r^2=(x+5)^2+(y-4)^2[/tex]
