Question

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

Answer 1

c)
`\sf r^2=(x+5)^2+(y-4)^2`

Explanation:

The center of the circle will be the midpoint of the endpoints.

`\sf midpoint=\left((x_1+x_2)/(2),(y_1+y_2)/(2)\right)`

Given:

• 
  `\sf (x_1,y_1)=(-8,2)`
• 
  `\sf (x_2,y_2)=(-2,6)`

`\sf \implies midpoint=center=\left((-8-2)/(2),(2+6)/(2)\right)=(-5,4)`

Equation of a circle:
`\sf (x-a)^2+(y-b)^2=r^2`

(where (a, b) is the center and r is the radius)

Substituting center (-5, 4) into the equation:

`\sf \implies (x-(-5))^2+(y-4)^2=r^2`

`\sf \implies (x+5)^2+(y-4)^2=r^2`

Therefore the solution is C:
`\sf r^2=(x+5)^2+(y-4)^2`