MATH SOLVE

3 months ago

Q:
# Help ASAP, 90 POINTS to anyone who answers it!!!A circle has a diameter with endpoints (-8, 2) and (-2, 6). What is the equation of the circle? r^2 = (x - 3)^2 + (y + 4)^2 r^2 = (x - 5)^2 + (y + 4)^2 r^2 = (x + 5)^2 + (y - 4)^2 r^2 = (x + 3)^2 + (y - 4)^2

Accepted Solution

A:

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 (x_1,y_1)=(-8,2)[/tex][tex]\sf (x_2,y_2)=(-2,6)[/tex][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]