Question

If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle? A) (x + 6)2 + (y + 10)2 = 20 B) (x − 6)2 + (y − 10)2 = 20 C) (x + 6)2 + (y + 10)2 = 2 5 D) (x − 6)2 + (y − 10)2 = 2 5

Answer 1

It's A...Just had it but i Chose B but it's A

Explanation:

Answer 2

`\large\boxed{A.\ (x+6)^2+(y+10)^2=20}`

Explanation:

The standard form of an equation of a circle:

`(x-h)^2+(y-k)^2=r^2`

(h, k) - center

r - radius

We have the endpoints of the diameter of a circle (-8, -6) and (-4, -14).

The midpoint of a diameter is a center of a circle.

The formula of a midpoint:

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

Substitute:

`x=(-8+(-4))/(2)=(-12)/(2)=-6\\\\y=(-6+(-14))/(2)=(-20)/(2)=-10`

We have h = -6 and k = -10.

The radius is the distance between a center and the point on a circumference of a circle.

The formula of a distance between two points:

`d=√((x_2-x_1)^2+(y_2-y_1)^2)`

Substitute (-6, -10) and (-8, -6):

`r=√((-8-(-6))^2+(-6-(-10))^2)=√((-2)^2+4^2)=√(4+16)=√(20)`

Finally we have

`(x-(-6))^2+(y-(-10))^2=(√(20))^2\\\\(x+6)^2+(y+10)^2=20`